3r^2+36r=12

Solution for 3r^2+36r=12 equation:

3r^2+36r=12

We move all terms to the left:

3r^2+36r-(12)=0

a = 3; b = 36; c = -12;
Δ = b2-4ac
Δ = 362-4·3·(-12)
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12\sqrt{10}}{2*3}=\frac{-36-12\sqrt{10}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12\sqrt{10}}{2*3}=\frac{-36+12\sqrt{10}}{6}
$