(X+2)(x-4)=(x+3)2

Solution for (X+2)(x-4)=(x+3)2 equation:

(X+2)(X-4)=(X+3)2

We move all terms to the left:

(X+2)(X-4)-((X+3)2)=0

We multiply parentheses ..

(+X^2-4X+2X-8)-((X+3)2)=0


We calculate terms in parentheses: -((X+3)2), so:

(X+3)2

We multiply parentheses

2X+6

Back to the equation:

-(2X+6)


We get rid of parentheses

X^2-4X+2X-2X-8-6=0

We add all the numbers together, and all the variables

X^2-4X-14=0

a = 1; b = -4; c = -14;
Δ = b2-4ac
Δ = -42-4·1·(-14)
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{2}}{2*1}=\frac{4-6\sqrt{2}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{2}}{2*1}=\frac{4+6\sqrt{2}}{2}
$