2x(x+4)-x(3x-2)=3x+8-4x+2x=1x+8

Solution for 2x(x+4)-x(3x-2)=3x+8-4x+2x=1x+8 equation:

2x(x+4)-x(3x-2)=3x+8-4x+2x=1x+8

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+9x-8=0

a = -1; b = 9; c = -8;
Δ = b2-4ac
Δ = 92-4·(-1)·(-8)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*-1}=\frac{-2}{-2}
=1
$