20x=0.15x^2

Solution for 20x=0.15x^2 equation:

20x=0.15x^2

We move all terms to the left:

20x-(0.15x^2)=0

We get rid of parentheses

-0.15x^2+20x=0

a = -0.15; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-0.15)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-0.15}=\frac{-40}{-0.3}
=133+0.1/0.3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-0.15}=\frac{0}{-0.3}
=0
$