x*(2x-6)=2*(4-3x-x^2)

Solution for x*(2x-6)=2*(4-3x-x^2) equation:

x(2x-6)=2(4-3x-x^2)

We move all terms to the left:

x(2x-6)-(2(4-3x-x^2))=0

We multiply parentheses

-(2(4-3x-x^2))+2x^2-6x=0


We calculate terms in parentheses: -(2(4-3x-x^2)), so:

2(4-3x-x^2)

We multiply parentheses

-2x^2-6x+8

Back to the equation:

-(-2x^2-6x+8)


We add all the numbers together, and all the variables

2x^2-(-2x^2-6x+8)-6x=0

We get rid of parentheses

2x^2+2x^2+6x-6x-8=0

We add all the numbers together, and all the variables

4x^2-8=0

a = 4; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·4·(-8)
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*4}=\frac{0-8\sqrt{2}}{8}
=-\frac{8\sqrt{2}}{8}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*4}=\frac{0+8\sqrt{2}}{8}
=\frac{8\sqrt{2}}{8}
=\sqrt{2}
$