(x2-2x)2-8=7(x2+2x)

Solution for (x2-2x)2-8=7(x2+2x) equation:

(x2-2x)2-8=7(x2+2x)

We move all terms to the left:

(x2-2x)2-8-(7(x2+2x))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(7(+x^2+2x)), so:

7(+x^2+2x)

We multiply parentheses

7x^2+14x

Back to the equation:

-(7x^2+14x)


We add all the numbers together, and all the variables

2x^2-4x-(7x^2+14x)-8=0

We get rid of parentheses

2x^2-7x^2-4x-14x-8=0

We add all the numbers together, and all the variables

-5x^2-18x-8=0

a = -5; b = -18; c = -8;
Δ = b2-4ac
Δ = -182-4·(-5)·(-8)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{41}}{2*-5}=\frac{18-2\sqrt{41}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{41}}{2*-5}=\frac{18+2\sqrt{41}}{-10}
$