(X-2)(x-2)-(x+1)(x-1)=5

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

(X-2)(X-2)-(X+1)(X-1)=5

We move all terms to the left:

(X-2)(X-2)-(X+1)(X-1)-(5)=0

We use the square of the difference formula

X^2+(X-2)(X-2)+1-5=0

We multiply parentheses ..

X^2+(+X^2-2X-2X+4)+1-5=0

We add all the numbers together, and all the variables

X^2+(+X^2-2X-2X+4)-4=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2X^2-4X=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$