-0.1x^2+40x-80=0

Solution for -0.1x^2+40x-80=0 equation:

-0.1x^2+40x-80=0

a = -0.1; b = 40; c = -80;
Δ = b2-4ac
Δ = 402-4·(-0.1)·(-80)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-28\sqrt{2}}{2*-0.1}=\frac{-40-28\sqrt{2}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+28\sqrt{2}}{2*-0.1}=\frac{-40+28\sqrt{2}}{-0.2}
$