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

Solution for x^2(3x+8)(3x-8)=0 equation:

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

We use the square of the difference formula

9x^2-64=0

a = 9; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·9·(-64)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*9}=\frac{-48}{18}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*9}=\frac{48}{18}
=2+2/3
$