36r^2+60r-15=0

Solution for 36r^2+60r-15=0 equation:

36r^2+60r-15=0

a = 36; b = 60; c = -15;
Δ = b2-4ac
Δ = 602-4·36·(-15)
Δ = 5760
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:

$\sqrt{\Delta}=\sqrt{5760}=\sqrt{576*10}=\sqrt{576}*\sqrt{10}=24\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-24\sqrt{10}}{2*36}=\frac{-60-24\sqrt{10}}{72}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+24\sqrt{10}}{2*36}=\frac{-60+24\sqrt{10}}{72}
$