(X+4)(x-4)+15=x+5

Solution for (X+4)(x-4)+15=x+5 equation:

(X+4)(X-4)+15=X+5

We move all terms to the left:

(X+4)(X-4)+15-(X+5)=0

We use the square of the difference formula

X^2-(X+5)-16+15=0

We get rid of parentheses

X^2-X-5-16+15=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
$