How do you solve using the quadratic formula 3x^2 + 6x = 12?

Mar 14, 2018

$- 1 \pm \sqrt{5}$

Explanation:

First get the 12 to the other side.
$3 {x}^{2} + 6 x - 12 = 0$

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{\text{2a}}$

where
a=3
b=6
c=-12

Now we just plug it in
$\frac{- \left(6\right) \pm \sqrt{{6}^{2} - 4 \left(3\right) \left(- 12\right)}}{\text{2(3)}}$

You can split this up like so
$- \frac{6}{\text{2(3)" +-sqrt(6^2-4(3)(-12))/"2(3)}}$

Then get this
$- \frac{6}{\text{6" +- sqrt(36-(-144))/"6}}$

Simplify
$- 1 \pm \frac{\sqrt{180}}{\text{6}}$

Then we simplify the square-root like so,
$\sqrt{36} \cdot \sqrt{5} = \sqrt{180}$

Thus
$6 \sqrt{5} = \sqrt{180}$

$- 1 \pm \frac{6 \sqrt{5}}{\text{6}}$

Which simplifies to
$- 1 \pm \sqrt{5}$