X^2+(x-8)=(x+8)

Solution for X^2+(x-8)=(x+8) equation:

X^2+(X-8)=(X+8)

We move all terms to the left:

X^2+(X-8)-((X+8))=0

We get rid of parentheses

X^2+X-((X+8))-8=0


We calculate terms in parentheses: -((X+8)), so:

(X+8)

We get rid of parentheses

X+8

Back to the equation:

-(X+8)


We get rid of parentheses

X^2+X-X-8-8=0

We add all the numbers together, and all the variables

X^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$