200q-8q^2=500+20q

Solution for 200q-8q^2=500+20q equation:

200q-8q^2=500+20q

We move all terms to the left:

200q-8q^2-(500+20q)=0

We add all the numbers together, and all the variables

-8q^2+200q-(20q+500)=0

We get rid of parentheses

-8q^2+200q-20q-500=0

We add all the numbers together, and all the variables

-8q^2+180q-500=0

a = -8; b = 180; c = -500;
Δ = b2-4ac
Δ = 1802-4·(-8)·(-500)
Δ = 16400
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:

$\sqrt{\Delta}=\sqrt{16400}=\sqrt{400*41}=\sqrt{400}*\sqrt{41}=20\sqrt{41}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-20\sqrt{41}}{2*-8}=\frac{-180-20\sqrt{41}}{-16}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+20\sqrt{41}}{2*-8}=\frac{-180+20\sqrt{41}}{-16}
$