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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-1x-3x-4=0

We add all the numbers together, and all the variables

2x^2-4x-4=0

a = 2; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·2·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*2}=\frac{4-4\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*2}=\frac{4+4\sqrt{3}}{4}
$