(4x-2)2=(2x-5)(8x+4)

Solution for (4x-2)2=(2x-5)(8x+4) equation:

(4x-2)2=(2x-5)(8x+4)

We move all terms to the left:

(4x-2)2-((2x-5)(8x+4))=0

We multiply parentheses

8x-((2x-5)(8x+4))-4=0

We multiply parentheses ..

-((+16x^2+8x-40x-20))+8x-4=0


We calculate terms in parentheses: -((+16x^2+8x-40x-20)), so:

(+16x^2+8x-40x-20)

We get rid of parentheses

16x^2+8x-40x-20

We add all the numbers together, and all the variables

16x^2-32x-20

Back to the equation:

-(16x^2-32x-20)


We add all the numbers together, and all the variables

8x-(16x^2-32x-20)-4=0

We get rid of parentheses

-16x^2+8x+32x+20-4=0

We add all the numbers together, and all the variables

-16x^2+40x+16=0

a = -16; b = 40; c = +16;
Δ = b2-4ac
Δ = 402-4·(-16)·16
Δ = 2624
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:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{41}}{2*-16}=\frac{-40-8\sqrt{41}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{41}}{2*-16}=\frac{-40+8\sqrt{41}}{-32}
$