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

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

(X-2)(X+3)+(X-2)(2X+1)+X^2-4=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
$