(2x-8)(x-4)=3

Solution for (2x-8)(x-4)=3 equation:

(2x-8)(x-4)=3

We move all terms to the left:

(2x-8)(x-4)-(3)=0

We multiply parentheses ..

(+2x^2-8x-8x+32)-3=0

We get rid of parentheses

2x^2-8x-8x+32-3=0

We add all the numbers together, and all the variables

2x^2-16x+29=0

a = 2; b = -16; c = +29;
Δ = b2-4ac
Δ = -162-4·2·29
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{6}}{2*2}=\frac{16-2\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{6}}{2*2}=\frac{16+2\sqrt{6}}{4}
$