(X+4)(x-5)=(x+2)(x-2)

Solution for (X+4)(x-5)=(x+2)(x-2) equation:

(X+4)(X-5)=(X+2)(X-2)

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

X^2+(+X^2-5X+4X-20)+4=0

We get rid of parentheses

X^2+X^2-5X+4X-20+4=0

We add all the numbers together, and all the variables

2X^2-1X-16=0

a = 2; b = -1; c = -16;
Δ = b2-4ac
Δ = -12-4·2·(-16)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2*2}=\frac{1-\sqrt{129}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*2}=\frac{1+\sqrt{129}}{4}
$