r2+20r+36=0

Solution for r2+20r+36=0 equation:

r2+20r+36=0

We add all the numbers together, and all the variables

r^2+20r+36=0

a = 1; b = 20; c = +36;
Δ = b2-4ac
Δ = 202-4·1·36
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*1}=\frac{-36}{2}
=-18
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*1}=\frac{-4}{2}
=-2
$