q^2+11q+10=-8

Solution for q^2+11q+10=-8 equation:

q^2+11q+10=-8

We move all terms to the left:

q^2+11q+10-(-8)=0

We add all the numbers together, and all the variables

q^2+11q+18=0

a = 1; b = 11; c = +18;
Δ = b2-4ac
Δ = 112-4·1·18
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*1}=\frac{-18}{2}
=-9
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*1}=\frac{-4}{2}
=-2
$