q=100+36q-4q^2

Solution for q=100+36q-4q^2 equation:

q=100+36q-4q^2

We move all terms to the left:

q-(100+36q-4q^2)=0

We get rid of parentheses

4q^2-36q+q-100=0

We add all the numbers together, and all the variables

4q^2-35q-100=0

a = 4; b = -35; c = -100;
Δ = b2-4ac
Δ = -352-4·4·(-100)
Δ = 2825
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:

$\sqrt{\Delta}=\sqrt{2825}=\sqrt{25*113}=\sqrt{25}*\sqrt{113}=5\sqrt{113}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{113}}{2*4}=\frac{35-5\sqrt{113}}{8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{113}}{2*4}=\frac{35+5\sqrt{113}}{8}
$