X=4x^2+2x^-4

Solution for X=4x^2+2x^-4 equation:

X=4X^2+2X^-4

We move all terms to the left:

X-(4X^2+2X^-4)=0

We get rid of parentheses

-4X^2+X-2X^+4=0

We add all the numbers together, and all the variables

-4X^2-1X+4=0

a = -4; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·(-4)·4
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{65}}{2*-4}=\frac{1-\sqrt{65}}{-8}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*-4}=\frac{1+\sqrt{65}}{-8}
$