q^2-20q+36=0

Solution for q^2-20q+36=0 equation:

q^2-20q+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:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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