(x-4)(x+4)=22x-8

Solution for (x-4)(x+4)=22x-8 equation:

(x-4)(x+4)=22x-8

We move all terms to the left:

(x-4)(x+4)-(22x-8)=0

We use the square of the difference formula

x^2-(22x-8)-16=0

We get rid of parentheses

x^2-22x+8-16=0

We add all the numbers together, and all the variables

x^2-22x-8=0

a = 1; b = -22; c = -8;
Δ = b2-4ac
Δ = -222-4·1·(-8)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{129}}{2*1}=\frac{22-2\sqrt{129}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{129}}{2*1}=\frac{22+2\sqrt{129}}{2}
$