X^2-3x+3x-2+x+2+x+5=40

Solution for X^2-3x+3x-2+x+2+x+5=40 equation:

X^2-3X+3X-2+X+2+X+5=40

We move all terms to the left:

X^2-3X+3X-2+X+2+X+5-(40)=0

We add all the numbers together, and all the variables

X^2+2X-35=0

a = 1; b = 2; c = -35;
Δ = b2-4ac
Δ = 22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*1}=\frac{-14}{2}
=-7
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*1}=\frac{10}{2}
=5
$