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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-8=0

a = 2; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·2·(-8)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*2}=\frac{8}{4}
=2
$