(2x-2)(4x+8)=(7x-1)(x+1)

Solution for (2x-2)(4x+8)=(7x-1)(x+1) equation:

(2x-2)(4x+8)=(7x-1)(x+1)

We move all terms to the left:

(2x-2)(4x+8)-((7x-1)(x+1))=0

We multiply parentheses ..

(+8x^2+16x-8x-16)-((7x-1)(x+1))=0


We calculate terms in parentheses: -((7x-1)(x+1)), so:

(7x-1)(x+1)

We multiply parentheses ..

(+7x^2+7x-1x-1)

We get rid of parentheses

7x^2+7x-1x-1

We add all the numbers together, and all the variables

7x^2+6x-1

Back to the equation:

-(7x^2+6x-1)


We get rid of parentheses

8x^2-7x^2+16x-8x-6x-16+1=0

We add all the numbers together, and all the variables

x^2+2x-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
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{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$