(5x+4)(5x-4)=32

Solution for (5x+4)(5x-4)=32 equation:

(5x+4)(5x-4)=32

We move all terms to the left:

(5x+4)(5x-4)-(32)=0

We use the square of the difference formula

25x^2-16-32=0

We add all the numbers together, and all the variables

25x^2-48=0

a = 25; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·25·(-48)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*25}=\frac{0-40\sqrt{3}}{50}
=-\frac{40\sqrt{3}}{50}
=-\frac{4\sqrt{3}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*25}=\frac{0+40\sqrt{3}}{50}
=\frac{40\sqrt{3}}{50}
=\frac{4\sqrt{3}}{5}
$