40x-0.1x^2=0

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

40x-0.1x^2=0

a = -0.1; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-0.1)·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*-0.1}=\frac{-80}{-0.2}
=+400
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-0.1}=\frac{0}{-0.2}
=0
$