40x=0.25x^2

Solution for 40x=0.25x^2 equation:

40x=0.25x^2

We move all terms to the left:

40x-(0.25x^2)=0

We get rid of parentheses

-0.25x^2+40x=0

a = -0.25; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-0.25)·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*-0.25}=\frac{-80}{-0.5}
=+160
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-0.25}=\frac{0}{-0.5}
=0
$