(x+3)(2x-1)-(x-2)(x+1)=39

Solution for (x+3)(2x-1)-(x-2)(x+1)=39 equation:

(x+3)(2x-1)-(x-2)(x+1)=39

We move all terms to the left:

(x+3)(2x-1)-(x-2)(x+1)-(39)=0

We multiply parentheses ..

(+2x^2-1x+6x-3)-(x-2)(x+1)-39=0

We get rid of parentheses

2x^2-1x+6x-(x-2)(x+1)-3-39=0

We multiply parentheses ..

2x^2-(+x^2+x-2x-2)-1x+6x-3-39=0

We add all the numbers together, and all the variables

2x^2-(+x^2+x-2x-2)+5x-42=0

We get rid of parentheses

2x^2-x^2-x+2x+5x+2-42=0

We add all the numbers together, and all the variables

x^2+6x-40=0

a = 1; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1·(-40)
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{-(6)-14}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*1}=\frac{8}{2}
=4
$