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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

x(3x+2)

We multiply parentheses

3x^2+2x

Back to the equation:

-(3x^2+2x)


We get rid of parentheses

4x^2-3x^2-2x-8=0

We add all the numbers together, and all the variables

x^2-2x-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$