q^2+21q+54=0

Solution for q^2+21q+54=0 equation:

q^2+21q+54=0

a = 1; b = 21; c = +54;
Δ = b2-4ac
Δ = 212-4·1·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-15}{2*1}=\frac{-36}{2}
=-18
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+15}{2*1}=\frac{-6}{2}
=-3
$