(x^2-6x)+(4x-14)=9x-42

Solution for (x^2-6x)+(4x-14)=9x-42 equation:

(x^2-6x)+(4x-14)=9x-42

We move all terms to the left:

(x^2-6x)+(4x-14)-(9x-42)=0

We get rid of parentheses

x^2-6x+4x-9x-14+42=0

We add all the numbers together, and all the variables

x^2-11x+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$