-x^2+20x=99

Solution for -x^2+20x=99 equation:

-x^2+20x=99

We move all terms to the left:

-x^2+20x-(99)=0

We add all the numbers together, and all the variables

-1x^2+20x-99=0

a = -1; b = 20; c = -99;
Δ = b2-4ac
Δ = 202-4·(-1)·(-99)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2}{2*-1}=\frac{-22}{-2}
=+11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2}{2*-1}=\frac{-18}{-2}
=+9
$