(2x-8)(2x+)=121

Solution for (2x-8)(2x+)=121 equation:

(2x-8)(2x+)=121

We move all terms to the left:

(2x-8)(2x+)-(121)=0

We add all the numbers together, and all the variables

(2x-8)(+2x)-121=0

We multiply parentheses ..

(+4x^2-16x)-121=0

We get rid of parentheses

4x^2-16x-121=0

a = 4; b = -16; c = -121;
Δ = b2-4ac
Δ = -162-4·4·(-121)
Δ = 2192
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:

$\sqrt{\Delta}=\sqrt{2192}=\sqrt{16*137}=\sqrt{16}*\sqrt{137}=4\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{137}}{2*4}=\frac{16-4\sqrt{137}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{137}}{2*4}=\frac{16+4\sqrt{137}}{8}
$