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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

2(2x-8)

We multiply parentheses

4x-16

Back to the equation:

-(4x-16)


We get rid of parentheses

-2x^2+4x-4x+16=0

We add all the numbers together, and all the variables

-2x^2+16=0

a = -2; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-2)·16
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-2}=\frac{0-8\sqrt{2}}{-4}
=-\frac{8\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-2}=\frac{0+8\sqrt{2}}{-4}
=\frac{8\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-1}
$