-w^2+20w-90=0

Solution for -w^2+20w-90=0 equation:

-w^2+20w-90=0

We add all the numbers together, and all the variables

-1w^2+20w-90=0

a = -1; b = 20; c = -90;
Δ = b2-4ac
Δ = 202-4·(-1)·(-90)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{10}}{2*-1}=\frac{-20-2\sqrt{10}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{10}}{2*-1}=\frac{-20+2\sqrt{10}}{-2}
$