(x+4)+(x+6)+(2x-3)+(x^2-2x)=42

Solution for (x+4)+(x+6)+(2x-3)+(x^2-2x)=42 equation:

(x+4)+(x+6)+(2x-3)+(x^2-2x)=42

We move all terms to the left:

(x+4)+(x+6)+(2x-3)+(x^2-2x)-(42)=0

We get rid of parentheses

x^2+x+x+2x-2x+4+6-3-42=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
$