(X^2-18x)-(4x^2-20x)=0

Solution for (X^2-18x)-(4x^2-20x)=0 equation:

(X^2-18X)-(4X^2-20X)=0

We get rid of parentheses

X^2-4X^2-18X+20X=0

We add all the numbers together, and all the variables

-3X^2+2X=0

a = -3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-3}=\frac{-4}{-6}
=2/3
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-3}=\frac{0}{-6}
=0
$