2x^2-22x^2+36x=0

Solution for 2x^2-22x^2+36x=0 equation:

2x^2-22x^2+36x=0

We add all the numbers together, and all the variables

-20x^2+36x=0

a = -20; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·(-20)·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*-20}=\frac{-72}{-40}
=1+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*-20}=\frac{0}{-40}
=0
$