(2x-1)(x+1)-(x-3)(x+5)=20

Solution for (2x-1)(x+1)-(x-3)(x+5)=20 equation:

(2x-1)(x+1)-(x-3)(x+5)=20

We move all terms to the left:

(2x-1)(x+1)-(x-3)(x+5)-(20)=0

We multiply parentheses ..

(+2x^2+2x-1x-1)-(x-3)(x+5)-20=0

We get rid of parentheses

2x^2+2x-1x-(x-3)(x+5)-1-20=0

We multiply parentheses ..

2x^2-(+x^2+5x-3x-15)+2x-1x-1-20=0

We add all the numbers together, and all the variables

2x^2-(+x^2+5x-3x-15)+x-21=0

We get rid of parentheses

2x^2-x^2-5x+3x+x+15-21=0

We add all the numbers together, and all the variables

x^2-1x-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$