2(3x+2)+2x(x-1)=4(x+2)

Solution for 2(3x+2)+2x(x-1)=4(x+2) equation:

2(3x+2)+2x(x-1)=4(x+2)

We move all terms to the left:

2(3x+2)+2x(x-1)-(4(x+2))=0

We multiply parentheses

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


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

4(x+2)

We multiply parentheses

4x+8

Back to the equation:

-(4x+8)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-4=0

a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4}
=-\frac{4\sqrt{2}}{4}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4}
=\frac{4\sqrt{2}}{4}
=\sqrt{2}
$