q^2-20q-108=0

Solution for q^2-20q-108=0 equation:

q^2-20q-108=0

a = 1; b = -20; c = -108;
Δ = b2-4ac
Δ = -202-4·1·(-108)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{13}}{2*1}=\frac{20-8\sqrt{13}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{13}}{2*1}=\frac{20+8\sqrt{13}}{2}
$