10=q^2+3q+2

Solution for 10=q^2+3q+2 equation:

10=q^2+3q+2

We move all terms to the left:

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

We get rid of parentheses

-q^2-3q-2+10=0

We add all the numbers together, and all the variables

-1q^2-3q+8=0

a = -1; b = -3; c = +8;
Δ = b2-4ac
Δ = -32-4·(-1)·8
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*-1}=\frac{3-\sqrt{41}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*-1}=\frac{3+\sqrt{41}}{-2}
$