q(32+q)=126

Solution for q(32+q)=126 equation:

q(32+q)=126

We move all terms to the left:

q(32+q)-(126)=0

We add all the numbers together, and all the variables

q(q+32)-126=0

We multiply parentheses

q^2+32q-126=0

a = 1; b = 32; c = -126;
Δ = b2-4ac
Δ = 322-4·1·(-126)
Δ = 1528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1528}=\sqrt{4*382}=\sqrt{4}*\sqrt{382}=2\sqrt{382}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{382}}{2*1}=\frac{-32-2\sqrt{382}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{382}}{2*1}=\frac{-32+2\sqrt{382}}{2}
$