2x^2=4x^-1

Solution for 2x^2=4x^-1 equation:

2x^2=4x^-1

We move all terms to the left:

2x^2-(4x^-1)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-4x+1=0

a = 2; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·2·1
Δ = 8
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:

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