x^2/((x-4)*(x+4))=16

Solution for x^2/((x-4)*(x+4))=16 equation:

x^2/((x-4)(x+4))=16

We move all terms to the left:

x^2/((x-4)(x+4))-(16)=0


Domain of the equation: ((x-4)(x+4))!=0

x∈R


We use the square of the difference formula

x^2-16-16=0

We add all the numbers together, and all the variables

x^2-32=0

a = 1; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·1·(-32)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*1}=\frac{0-8\sqrt{2}}{2}
=-\frac{8\sqrt{2}}{2}
=-4\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*1}=\frac{0+8\sqrt{2}}{2}
=\frac{8\sqrt{2}}{2}
=4\sqrt{2}
$