3r^2=12r+15

Solution for 3r^2=12r+15 equation:

3r^2=12r+15

We move all terms to the left:

3r^2-(12r+15)=0

We get rid of parentheses

3r^2-12r-15=0

a = 3; b = -12; c = -15;
Δ = b2-4ac
Δ = -122-4·3·(-15)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*3}=\frac{-6}{6}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*3}=\frac{30}{6}
=5
$