-0.8x^2+40x=0

Solution for -0.8x^2+40x=0 equation:

-0.8x^2+40x=0

a = -0.8; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-0.8)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-0.8}=\frac{-80}{-1.6}
=+50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-0.8}=\frac{0}{-1.6}
=0
$