2r^2+12r-12=-3

Solution for 2r^2+12r-12=-3 equation:

2r^2+12r-12=-3

We move all terms to the left:

2r^2+12r-12-(-3)=0

We add all the numbers together, and all the variables

2r^2+12r-9=0

a = 2; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·2·(-9)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{6}}{2*2}=\frac{-12-6\sqrt{6}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{6}}{2*2}=\frac{-12+6\sqrt{6}}{4}
$