0=100+36q-4q^2

Solution for 0=100+36q-4q^2 equation:

0=100+36q-4q^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4q^2-36q-100=0

a = 4; b = -36; c = -100;
Δ = b2-4ac
Δ = -362-4·4·(-100)
Δ = 2896
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:

$\sqrt{\Delta}=\sqrt{2896}=\sqrt{16*181}=\sqrt{16}*\sqrt{181}=4\sqrt{181}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{181}}{2*4}=\frac{36-4\sqrt{181}}{8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{181}}{2*4}=\frac{36+4\sqrt{181}}{8}
$