x+4+x+3+x^2-2x+2x-4=40

Solution for x+4+x+3+x^2-2x+2x-4=40 equation:

x+4+x+3+x^2-2x+2x-4=40

We move all terms to the left:

x+4+x+3+x^2-2x+2x-4-(40)=0

We add all the numbers together, and all the variables

x^2+2x-37=0

a = 1; b = 2; c = -37;
Δ = b2-4ac
Δ = 22-4·1·(-37)
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{38}}{2*1}=\frac{-2-2\sqrt{38}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{38}}{2*1}=\frac{-2+2\sqrt{38}}{2}
$