2n=2(2n)^2-5

Solution for 2n=2(2n)^2-5 equation:

2n=2(2n)^2-5

We move all terms to the left:

2n-(2(2n)^2-5)=0

We get rid of parentheses

-22n^2+2n+5=0

a = -22; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-22)·5
Δ = 444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{444}=\sqrt{4*111}=\sqrt{4}*\sqrt{111}=2\sqrt{111}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{111}}{2*-22}=\frac{-2-2\sqrt{111}}{-44}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{111}}{2*-22}=\frac{-2+2\sqrt{111}}{-44}
$