-2x^2+20x-40=0

Solution for -2x^2+20x-40=0 equation:

-2x^2+20x-40=0

a = -2; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·(-2)·(-40)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{5}}{2*-2}=\frac{-20-4\sqrt{5}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{5}}{2*-2}=\frac{-20+4\sqrt{5}}{-4}
$