9q^2+36q+27=0

Solution for 9q^2+36q+27=0 equation:

9q^2+36q+27=0

a = 9; b = 36; c = +27;
Δ = b2-4ac
Δ = 362-4·9·27
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-18}{2*9}=\frac{-54}{18}
=-3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+18}{2*9}=\frac{-18}{18}
=-1
$