Solve $3x^2+4x-5=0$: Easy 2-Decimal Place Guide

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Hey guys! Ever stumbled upon a quadratic equation and thought, "What fresh math hell is this?" Don't worry, we've all been there. Today, we're diving deep into solving the specific quadratic equation $3x^2+4x-5=0$, and trust me, by the end of this, you'll be nailing it to two decimal places like a pro. We're talking about a problem that pops up everywhere, from high school math classes to engineering blueprints, so understanding how to tackle it is super valuable. Let's break down this beast step-by-step, making sure we don't leave any math stone unturned. We'll be using the trusty quadratic formula, which is basically your best friend when factoring gets too tricky. Remember, the goal is to find the values of 'x' that make this equation true, and we want those answers rounded neatly to two decimal places. So, grab your calculators, maybe a comfy chair, and let's get this math party started! We'll cover the 'why' behind the formula and the 'how' of applying it, making sure it all clicks. Get ready to feel confident about solving quadratic equations, because after this, you'll be unstoppable.

Understanding the Quadratic Formula: Your Math Superpower

Alright, before we jump into solving our specific equation, $3x^2+4x-5=0$, let's chat about the hero of our story: the quadratic formula. You'll often see it looking like this: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Now, I know what you might be thinking, "Looks complicated!" But honestly, it's a lifesaver. This formula is derived from the general form of a quadratic equation, which is $ax^2 + bx + c = 0$. Here, 'a', 'b', and 'c' are just coefficients (those numbers sitting next to the x's or standing alone), and 'a' can't be zero, otherwise, it wouldn't be quadratic anymore, right? The beauty of this formula is that it works for any quadratic equation, no matter how ugly or simple. It's like a universal key that unlocks all quadratic solutions. The '±' symbol is key here, guys; it means we're going to get two potential answers for 'x', one using the plus sign and one using the minus sign. The part under the square root, $b^2-4ac$, is super important too. It's called the discriminant, and it tells us what kind of solutions we're going to get (real, imaginary, or just one repeated solution). For our equation, $3x^2+4x-5=0$, we need to identify our 'a', 'b', and 'c' values. Looking at it, it's pretty straightforward: 'a' is 3 (the coefficient of $x^2$), 'b' is 4 (the coefficient of x), and 'c' is -5 (the constant term). We'll plug these numbers into the formula, and the magic will happen. Understanding this formula isn't just about memorizing it; it's about appreciating its power to simplify complex problems. Think of it as your trusty sidekick in the world of algebra, always ready to help you find those elusive 'x' values. We're not just plugging and chugging; we're understanding the mechanics behind finding solutions, which makes the whole process much less daunting and way more rewarding.

Plugging in the Numbers: Solving $3x^2+4x-5=0$

Okay, team, it's time to get our hands dirty with our specific equation: $3x^2+4x-5=0$. We've already identified our players: $a=3$, $b=4$, and $c=-5$. Now, we're going to substitute these values carefully into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Let's start plugging:

$x = \frac{-(4) \pm \sqrt{(4)^2-4(3)(-5)}}{2(3)}$

See? Not so scary when you break it down. Now, let's simplify this step-by-step. First, let's tackle the stuff inside the square root (the discriminant):

$b^2-4ac = (4)^2 - 4(3)(-5)$

$b^2-4ac = 16 - (-60)$

$b^2-4ac = 16 + 60$

$b^2-4ac = 76$

Awesome! So, the square root part is $\sqrt{76}$. Now let's handle the denominator:

$2a = 2(3) = 6$

And the '-b' part is just -4.

So, our formula now looks like this:

$x = \frac{-4 \pm \sqrt{76}}{6}$

This is where our calculator comes in handy for getting those decimal places. We need to calculate $\sqrt{76}$.

$\sqrt{76} \approx 8.7177978...$

Now, we're going to split this into our two possible solutions, one with '+' and one with '-'.

Solution 1 (using '+'):

$x_1 = \frac{-4 + 8.7177978...}{6}$

$x_1 = \frac{4.7177978...}{6}$

$x_1 \approx 0.7862996...$

Solution 2 (using '-'):

$x_2 = \frac{-4 - 8.7177978...}{6}$

$x_2 = \frac{-12.7177978...}{6}$

$x_2 \approx -2.1196329...$

See? We've successfully plugged in the numbers and performed the calculations. It's all about being methodical. Don't rush these steps, and double-check your substitutions. Every number has its place, and understanding where each part of the formula comes from makes the process feel less like random number-crunching and more like logical problem-solving. We're getting closer to our final, rounded answers, and the end is in sight!

Rounding to Perfection: The Final Touch

We're in the home stretch, guys! We've done the heavy lifting of plugging values into the quadratic formula and calculating the intermediate steps for $3x^2+4x-5=0$. Our two raw answers are approximately $0.7862996...$ and $-2.1196329...$. The problem specifically asks for the answers to 2 decimal places. This is a crucial final step that often trips people up if they aren't careful. Rounding is all about looking at the digit after the last digit you want to keep.

Let's take our first answer: $x_1 \approx 0.7862996...$

We want to keep two decimal places, so we look at the first two digits: 0.78. The third decimal digit is 6. The rule is: if this digit is 5 or greater, we round up the second decimal digit. If it's less than 5, we keep the second decimal digit as it is. Since 6 is greater than 5, we round up the 8 to a 9.

So, $x_1$ rounded to two decimal places is $0.79$.

Now, let's do the same for our second answer: $x_2 \approx -2.1196329...$

We look at the first two decimal places: -2.11. The third decimal digit is 9. Again, 9 is greater than 5, so we need to round up the second decimal digit (the 1). Rounding up 1 gives us 2.

So, $x_2$ rounded to two decimal places is $-2.12$.

And there you have it! The solutions to the quadratic equation $3x^2+4x-5=0$, rounded to two decimal places, are approximately $0.79$ and $-2.12$. It's a neat feeling to take those messy, long decimal numbers and present them in a clean, usable format. This skill of rounding is vital not just in math but in pretty much any field where you deal with measurements or data. Always pay attention to the rounding instructions in a problem – it's the final polish that shows you've completed the task accurately. Practice this a few times, and you'll be rounding like a champ in no time. You've conquered this quadratic equation, guys – pat yourselves on the back!

Checking Your Work: Ensuring Accuracy

Now, for the ultimate satisfaction check – making sure our answers are actually correct for $3x^2+4x-5=0$. We found our solutions to be approximately $x_1 \approx 0.79$ and $x_2 \approx -2.12$. The best way to check is to plug these values back into the original equation and see if we get something close to zero. Remember, because we rounded, we won't get exactly zero, but it should be very, very close.

Checking $x_1 \approx 0.79$:

Let's substitute $0.79$ for 'x' in $3x^2+4x-5=0$:

$3(0.79)^2 + 4(0.79) - 5$

First, calculate $(0.79)^2$: $0.79 \times 0.79 = 0.6241$

Now, substitute that back:

$3(0.6241) + 4(0.79) - 5$

Perform the multiplications:

$1.8723 + 3.16 - 5$

Now, add and subtract:

$5.0323 - 5 = 0.0323$

This is very close to zero! The small difference is due to the rounding we did earlier. If we had used the more precise value of $x_1 \approx 0.786...$, we would get much closer to zero.

Checking $x_2 \approx -2.12$:

Now, let's substitute $-2.12$ for 'x' in $3x^2+4x-5=0$:

$3(-2.12)^2 + 4(-2.12) - 5$

First, calculate $(-2.12)^2$: $(-2.12) \times (-2.12) = 4.4944$

Now, substitute that back:

$3(4.4944) + 4(-2.12) - 5$

Perform the multiplications:

$13.4832 - 8.48 - 5$

Now, add and subtract:

$5.0032 - 5 = 0.0032$

Again, this is extremely close to zero! This gives us a high level of confidence that our rounded answers are correct. This checking step is super important, guys. It prevents silly errors and ensures you've truly mastered the problem. If you get a number far from zero, it's a sign to go back and review your calculations, especially the substitution and the arithmetic steps. It's all part of the learning process!

Conclusion: You've Got This!

So there you have it! We've successfully navigated the world of solving the quadratic equation $3x^2+4x-5=0$ and provided the answers rounded to two decimal places. We've broken down the mighty quadratic formula, meticulously plugged in our values for 'a', 'b', and 'c', performed the calculations, and finally, rounded our results to perfection. The solutions are approximately $x = 0.79$ and $x = -2.12$. Remember, the quadratic formula is your go-to tool for any equation in the form $ax^2+bx+c=0$ when factoring becomes a headache. The key is to identify 'a', 'b', and 'c' correctly, substitute them carefully, and then work through the calculations step-by-step, paying close attention to the order of operations and the discriminant ($b^2-4ac$). Don't forget the final, crucial step of rounding to the specified number of decimal places, and always, always check your work by plugging your answers back into the original equation. Math can seem intimidating, but by breaking down problems into smaller, manageable steps and understanding the logic behind the tools we use, even complex equations become solvable. You guys have tackled a real math challenge today, and hopefully, you feel more confident and empowered. Keep practicing, keep exploring, and never shy away from a good math problem! You've totally got this!