(8-3x)(8+3x)=2

Solution for (8-3x)(8+3x)=2 equation:

(8-3x)(8+3x)=2

We move all terms to the left:

(8-3x)(8+3x)-(2)=0

We add all the numbers together, and all the variables

(-3x+8)(3x+8)-2=0

We multiply parentheses ..

(-9x^2-24x+24x+64)-2=0

We get rid of parentheses

-9x^2-24x+24x+64-2=0

We add all the numbers together, and all the variables

-9x^2+62=0

a = -9; b = 0; c = +62;
Δ = b2-4ac
Δ = 02-4·(-9)·62
Δ = 2232
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:

$\sqrt{\Delta}=\sqrt{2232}=\sqrt{36*62}=\sqrt{36}*\sqrt{62}=6\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{62}}{2*-9}=\frac{0-6\sqrt{62}}{-18}
=-\frac{6\sqrt{62}}{-18}
=-\frac{\sqrt{62}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{62}}{2*-9}=\frac{0+6\sqrt{62}}{-18}
=\frac{6\sqrt{62}}{-18}
=\frac{\sqrt{62}}{-3}
$