(2a-4)(2a+4)=8a2-64

Solution for (2a-4)(2a+4)=8a2-64 equation:

(2a-4)(2a+4)=8a^2-64

We move all terms to the left:

(2a-4)(2a+4)-(8a^2-64)=0

We use the square of the difference formula

4a^2-(8a^2-64)-16=0

We get rid of parentheses

4a^2-8a^2+64-16=0

We add all the numbers together, and all the variables

-4a^2+48=0

a = -4; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-4)·48
Δ = 768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*-4}=\frac{0-16\sqrt{3}}{-8}
=-\frac{16\sqrt{3}}{-8}
=-\frac{2\sqrt{3}}{-1}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*-4}=\frac{0+16\sqrt{3}}{-8}
=\frac{16\sqrt{3}}{-8}
=\frac{2\sqrt{3}}{-1}
$