(x-4)(x+2)+3(x-2);x=3

Solution for (x-4)(x+2)+3(x-2);x=3 equation:

(x-4)(x+2)+3(x-2)x=3

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

3x^2+(+x^2+2x-4x-8)-6x-3=0

We get rid of parentheses

3x^2+x^2+2x-4x-6x-8-3=0

We add all the numbers together, and all the variables

4x^2-8x-11=0

a = 4; b = -8; c = -11;
Δ = b2-4ac
Δ = -82-4·4·(-11)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{15}}{2*4}=\frac{8-4\sqrt{15}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{15}}{2*4}=\frac{8+4\sqrt{15}}{8}
$