# How do you use the quadratic formula to find both solutions to the quadratic equation 3x^2 + 6x = 12?

Apr 27, 2018

$x = - 1 \pm \sqrt{5}$

Here's how I did it:

#### Explanation:

$3 {x}^{2} + 6 x = 12$

To use the quadratic formula, we first have to set the quadratic equal to zero.

To do this, let's subtract $\textcolor{red}{12}$ from both sides of the equation:
$3 {x}^{2} + 6 x \quad \textcolor{red}{- \quad 12} = 12 \quad \textcolor{red}{- \quad 12}$

$3 {x}^{2} + 6 x - 12 = 0$

Now, the equation is in standard form, or $\textcolor{red}{a} {x}^{2} + \textcolor{m a \ge n t a}{b} x + \textcolor{b l u e}{c}$, so:
$\textcolor{red}{a = 3}$, $\textcolor{m a \ge n t a}{b = 6}$, and $\textcolor{b l u e}{c = - 12}$

The quadratic formula can solve for $x$ in any case, and we use it especially when the expression is not factorable.
$x = \frac{\textcolor{m a \ge n t a}{- b} \pm \sqrt{{\textcolor{m a \ge n t a}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{b l u e}{c}}}{2 \textcolor{red}{a}}$

Now we can plug in our values for $a$, $b$, and $c$:
$x = \frac{\textcolor{m a \ge n t a}{- 6} \pm \sqrt{{\left(\textcolor{m a \ge n t a}{- 6}\right)}^{2} - 4 \left(\textcolor{red}{3}\right) \left(\textcolor{b l u e}{- 12}\right)}}{2 \left(\textcolor{red}{3}\right)}$

Simplify:
$x = \frac{- 6 \pm \sqrt{36 - 4 \left(- 36\right)}}{6}$

$x = \frac{- 6 \pm \sqrt{36 + 144}}{6}$

$x = \frac{- 6 \pm \sqrt{180}}{6}$

$x = \frac{- 6 \pm 6 \sqrt{5}}{6}$

Divide by the denominator to get the simplified answer:
$x = - 1 \pm \sqrt{5}$

Hope this helps!