-q^2+200q-100=0

Solution for -q^2+200q-100=0 equation:

-q^2+200q-100=0

We add all the numbers together, and all the variables

-1q^2+200q-100=0

a = -1; b = 200; c = -100;
Δ = b2-4ac
Δ = 2002-4·(-1)·(-100)
Δ = 39600
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:

$\sqrt{\Delta}=\sqrt{39600}=\sqrt{3600*11}=\sqrt{3600}*\sqrt{11}=60\sqrt{11}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-60\sqrt{11}}{2*-1}=\frac{-200-60\sqrt{11}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+60\sqrt{11}}{2*-1}=\frac{-200+60\sqrt{11}}{-2}
$