2x(3x-4)=(2x2+8)

Solution for 2x(3x-4)=(2x2+8) equation:

2x(3x-4)=(2x^2+8)

We move all terms to the left:

2x(3x-4)-((2x^2+8))=0

We multiply parentheses

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


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

(2x^2+8)

We get rid of parentheses

2x^2+8

Back to the equation:

-(2x^2+8)


We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-8x-8=0

a = 4; b = -8; c = -8;
Δ = b2-4ac
Δ = -82-4·4·(-8)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{3}}{2*4}=\frac{8-8\sqrt{3}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{3}}{2*4}=\frac{8+8\sqrt{3}}{8}
$