How do you solve 5m^2 - 11m - 3 = 0 using the quadratic formula?

Oct 16, 2015

First, we determine the values of a, b, and c that go into the quadratic formula, then plug them in and solve...

Explanation:

So, when we have a quadratic equation:

$a {x}^{2} + b x + c = 0$

One of the ways we can solve for $x$ is by using the quadratic formula:

But wait... for your problem, there's no $x$! Taking a look at the equation, the first thing we need to do is realize that $x$ is just a placeholder for any variable (a number that we don't know... yet). In your question, the variable is shown as $m$. It's the same thing; we could call it $x$ or $m$ or $\omega$ or whatever we like.

Now, we need to figure out the values of $a$, $b$, and $c$. $a$ is the number before the ${x}^{2}$ variable (or, in your case, ${m}^{2}$). $b$ is the number before $m$, and c is the number not associated with the variable. So...

$a = 5$

$b = - 11$ (don't forget that negative sign!)

$c = - 3$ (again, the negative sign is important)

Now, we can take those three numbers and plug them into the quadratic equation:

$x = \left(- \left(- 11\right) \left(\pm\right) \frac{\sqrt{\left(- {11}^{2}\right) - \left(4\right) \left(5\right) \left(- 3\right)}}{\left(2\right) \left(5\right)}\right)$

See that $\pm$ in there? That means we're going to get two answers.

Either

$x = \left(- \left(- 11\right) \left(+\right) \frac{\sqrt{\left(- {11}^{2}\right) - \left(4\right) \left(5\right) \left(- 3\right)}}{\left(2\right) \left(5\right)}\right)$

OR

$x = \left(- \left(- 11\right) \left(-\right) \frac{\sqrt{\left(- {11}^{2}\right) - \left(4\right) \left(5\right) \left(- 3\right)}}{\left(2\right) \left(5\right)}\right)$

$x$ can either be 2.445 or -0.2453 (leaving off quite a few decimal places)

Let's check to make sure that's right, by plugging in our values for $x$:

Does $\left(5\right) \left({2.445}^{2}\right) + \left(- 11\right) \left(2.445\right) + \left(- 3\right) = 0$?

Does $\left(5\right) \left({0.2453}^{2}\right) + \left(- 11\right) \left(0.2453\right) + \left(- 3\right) = 0$?