(x+2)*(2x-3)+x-4=6

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

x+(x+2)(2x-3)-10=0

We multiply parentheses ..

(+2x^2-3x+4x-6)+x-10=0

We get rid of parentheses

2x^2-3x+4x+x-6-10=0

We add all the numbers together, and all the variables

2x^2+2x-16=0

a = 2; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·2·(-16)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{33}}{2*2}=\frac{-2-2\sqrt{33}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{33}}{2*2}=\frac{-2+2\sqrt{33}}{4}
$