r^2+20r-36=-7

Solution for r^2+20r-36=-7 equation:

r^2+20r-36=-7

We move all terms to the left:

r^2+20r-36-(-7)=0

We add all the numbers together, and all the variables

r^2+20r-29=0

a = 1; b = 20; c = -29;
Δ = b2-4ac
Δ = 202-4·1·(-29)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{129}}{2*1}=\frac{-20-2\sqrt{129}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{129}}{2*1}=\frac{-20+2\sqrt{129}}{2}
$