3q+q^2=-2

Solution for 3q+q^2=-2 equation:

3q+q^2=-2

We move all terms to the left:

3q+q^2-(-2)=0

We add all the numbers together, and all the variables

q^2+3q+2=0

a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*1}=\frac{-4}{2}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*1}=\frac{-2}{2}
=-1
$