(X+6)(x-4)+(x+2)(x-2)=x^2-4

Solution for (X+6)(x-4)+(x+2)(x-2)=x^2-4 equation:

(X+6)(X-4)+(X+2)(X-2)=X^2-4

We move all terms to the left:

(X+6)(X-4)+(X+2)(X-2)-(X^2-4)=0

We use the square of the difference formula

X^2+(X+6)(X-4)-(X^2-4)-4=0

We get rid of parentheses

X^2-X^2+(X+6)(X-4)+4-4=0

We multiply parentheses ..

X^2-X^2+(+X^2-4X+6X-24)+4-4=0

We add all the numbers together, and all the variables

(+X^2-4X+6X-24)=0

We get rid of parentheses

X^2-4X+6X-24=0

We add all the numbers together, and all the variables

X^2+2X-24=0

a = 1; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·1·(-24)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*1}=\frac{-12}{2}
=-6
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*1}=\frac{8}{2}
=4
$