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

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

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

We move all terms to the left:

2x+2-((x-2)(x+4))=0

We multiply parentheses ..

-((+x^2+4x-2x-8))+2x+2=0


We calculate terms in parentheses: -((+x^2+4x-2x-8)), so:

(+x^2+4x-2x-8)

We get rid of parentheses

x^2+4x-2x-8

We add all the numbers together, and all the variables

x^2+2x-8

Back to the equation:

-(x^2+2x-8)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2+2x-2x+8+2=0

We add all the numbers together, and all the variables

-1x^2+10=0

a = -1; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-1)·10
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*-1}=\frac{0-2\sqrt{10}}{-2}
=-\frac{2\sqrt{10}}{-2}
=-\frac{\sqrt{10}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-1}=\frac{0+2\sqrt{10}}{-2}
=\frac{2\sqrt{10}}{-2}
=\frac{\sqrt{10}}{-1}
$