2x^2-20x=x^2-19

Solution for 2x^2-20x=x^2-19 equation:

2x^2-20x=x^2-19

We move all terms to the left:

2x^2-20x-(x^2-19)=0

We get rid of parentheses

2x^2-x^2-20x+19=0

We add all the numbers together, and all the variables

x^2-20x+19=0

a = 1; b = -20; c = +19;
Δ = b2-4ac
Δ = -202-4·1·19
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-18}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+18}{2*1}=\frac{38}{2}
=19
$