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

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Answer 1 · 2018-03-14T19:54:52

$-1 +- sqrt5$

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

Quadratic Formula is
$(-b +- sqrt(b^2-4ac))/"2a"$

where
a=3
b=6
c=-12

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

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

Then get this
$-6/"6" +- sqrt(36-(-144))/"6"$

Simplify
$-1 +- sqrt(180)/"6"$

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

Thus
$6sqrt(5)=sqrt(180)$

$-1 +-(6sqrt(5))/"6"$

Which simplifies to
$-1 +-sqrt(5)$

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