3r^2+27r=12

Solution for 3r^2+27r=12 equation:

3r^2+27r=12

We move all terms to the left:

3r^2+27r-(12)=0

a = 3; b = 27; c = -12;
Δ = b2-4ac
Δ = 272-4·3·(-12)
Δ = 873
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:

$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{97}}{2*3}=\frac{-27-3\sqrt{97}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{97}}{2*3}=\frac{-27+3\sqrt{97}}{6}
$