2^4-18x^2+16=0

Solution for 2^4-18x^2+16=0 equation:

2^4-18x^2+16=0

We add all the numbers together, and all the variables

-18x^2+32=0

a = -18; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-18)·32
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*-18}=\frac{-48}{-36}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-18}=\frac{48}{-36}
=-1+1/3
$