A(x+4)+B(-x-2)(x+4)+C(-x-2)=6x-18

Solution for A(x+4)+B(-x-2)(x+4)+C(-x-2)=6x-18 equation:

(A+4)+(-A-2)(A+4)+(-A-2)=6A-18

We move all terms to the left:

(A+4)+(-A-2)(A+4)+(-A-2)-(6A-18)=0

We add all the numbers together, and all the variables

(A+4)+(-1A-2)(A+4)+(-1A-2)-(6A-18)=0

We get rid of parentheses

A+(-1A-2)(A+4)-1A-6A+4-2+18=0

We multiply parentheses ..

(-1A^2-4A-2A-8)+A-1A-6A+4-2+18=0

We add all the numbers together, and all the variables

(-1A^2-4A-2A-8)-6A+20=0

We get rid of parentheses

-1A^2-4A-2A-6A-8+20=0

We add all the numbers together, and all the variables

-1A^2-12A+12=0

a = -1; b = -12; c = +12;
Δ = b2-4ac
Δ = -122-4·(-1)·12
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{3}}{2*-1}=\frac{12-8\sqrt{3}}{-2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{3}}{2*-1}=\frac{12+8\sqrt{3}}{-2}
$