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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(3(x+4)(x-3)-x(x-3)), so:

3(x+4)(x-3)-x(x-3)

We multiply parentheses

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

We multiply parentheses ..

-x^2+3(+x^2-3x+4x-12)+3x

We add all the numbers together, and all the variables

-1x^2+3(+x^2-3x+4x-12)+3x

We multiply parentheses

-1x^2+3x^2-9x+12x+3x-36

We add all the numbers together, and all the variables

2x^2+6x-36

Back to the equation:

-(2x^2+6x-36)


We get rid of parentheses

x^2-2x^2-1x+2x-3x-6x-2+36+2=0

We add all the numbers together, and all the variables

-1x^2-8x+36=0

a = -1; b = -8; c = +36;
Δ = b2-4ac
Δ = -82-4·(-1)·36
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{13}}{2*-1}=\frac{8-4\sqrt{13}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{13}}{2*-1}=\frac{8+4\sqrt{13}}{-2}
$