3r^2+6r-12=0

Solution for 3r^2+6r-12=0 equation:

3r^2+6r-12=0

a = 3; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·3·(-12)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{5}}{2*3}=\frac{-6-6\sqrt{5}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{5}}{2*3}=\frac{-6+6\sqrt{5}}{6}
$