-2x^2+160=40x

Solution for -2x^2+160=40x equation:

-2x^2+160=40x

We move all terms to the left:

-2x^2+160-(40x)=0

a = -2; b = -40; c = +160;
Δ = b2-4ac
Δ = -402-4·(-2)·160
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{5}}{2*-2}=\frac{40-24\sqrt{5}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{5}}{2*-2}=\frac{40+24\sqrt{5}}{-4}
$