(4x+12)(4x-12)=135

Solution for (4x+12)(4x-12)=135 equation:

(4x+12)(4x-12)=135

We move all terms to the left:

(4x+12)(4x-12)-(135)=0

We use the square of the difference formula

16x^2-144-135=0

We add all the numbers together, and all the variables

16x^2-279=0

a = 16; b = 0; c = -279;
Δ = b2-4ac
Δ = 02-4·16·(-279)
Δ = 17856
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:

$\sqrt{\Delta}=\sqrt{17856}=\sqrt{576*31}=\sqrt{576}*\sqrt{31}=24\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{31}}{2*16}=\frac{0-24\sqrt{31}}{32}
=-\frac{24\sqrt{31}}{32}
=-\frac{3\sqrt{31}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{31}}{2*16}=\frac{0+24\sqrt{31}}{32}
=\frac{24\sqrt{31}}{32}
=\frac{3\sqrt{31}}{4}
$