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

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Answer 1 · 2018-03-14T19:54:52

$- 1 \pm \sqrt{5}$ $-1 +- sqrt5$

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

Quadratic Formula is
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2a}$ $(-b +- sqrt(b^2-4ac))/"2a"$

where
a=3
b=6
c=-12

Now we just plug it in
$\frac{- (6) \pm \sqrt{6^{2} - 4 (3) (- 12)}}{2(3)}$ $(-(6)+-sqrt(6^2-4(3)(-12)))/"2(3)"$

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

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

Simplify
$- 1 \pm \frac{\sqrt{180}}{6}$ $-1 +- sqrt(180)/"6"$

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

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

$- 1 \pm \frac{6 \sqrt{5}}{6}$ $-1 +-(6sqrt(5))/"6"$

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

How do you solve using the quadratic formula 3x^2 + 6x = 123x2+6x=123x^2 + 6x = 12?