(x-2)(x+3)+(x-2)(2x+1)+x2-4=0

Solution for (x-2)(x+3)+(x-2)(2x+1)+x2-4=0 equation:

(x-2)(x+3)+(x-2)(2x+1)+x2-4=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

x^2+(+x^2+3x-2x-6)+(x-2)(2x+1)-4=0

We get rid of parentheses

x^2+x^2+3x-2x+(x-2)(2x+1)-6-4=0

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-2x-12=0

a = 4; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·4·(-12)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*4}=\frac{-12}{8}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*4}=\frac{16}{8}
=2
$