(r-3)(r+15)=-36

Solution for (r-3)(r+15)=-36 equation:

(r-3)(r+15)=-36

We move all terms to the left:

(r-3)(r+15)-(-36)=0

We add all the numbers together, and all the variables

(r-3)(r+15)+36=0

We multiply parentheses ..

(+r^2+15r-3r-45)+36=0

We get rid of parentheses

r^2+15r-3r-45+36=0

We add all the numbers together, and all the variables

r^2+12r-9=0

a = 1; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·1·(-9)
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{-(12)-6\sqrt{5}}{2*1}=\frac{-12-6\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{5}}{2*1}=\frac{-12+6\sqrt{5}}{2}
$