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

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

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We add all the numbers together, and all the variables

2x^2-3x-7=0

a = 2; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·2·(-7)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{65}}{2*2}=\frac{3-\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*2}=\frac{3+\sqrt{65}}{4}
$