4-8x2+6x=2x(x+3)+2x

Solution for 4-8x2+6x=2x(x+3)+2x equation:

4-8x^2+6x=2x(x+3)+2x

We move all terms to the left:

4-8x^2+6x-(2x(x+3)+2x)=0


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

2x(x+3)+2x

We add all the numbers together, and all the variables

2x+2x(x+3)

We multiply parentheses

2x^2+2x+6x

We add all the numbers together, and all the variables

2x^2+8x

Back to the equation:

-(2x^2+8x)


We get rid of parentheses

-8x^2-2x^2+6x-8x+4=0

We add all the numbers together, and all the variables

-10x^2-2x+4=0

a = -10; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-10)·4
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{41}}{2*-10}=\frac{2-2\sqrt{41}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{41}}{2*-10}=\frac{2+2\sqrt{41}}{-20}
$