(-4n^2+2n)-(n^2-5)=0

Solution for (-4n^2+2n)-(n^2-5)=0 equation:

(-4n^2+2n)-(n^2-5)=0

We get rid of parentheses

-4n^2-n^2+2n+5=0

We add all the numbers together, and all the variables

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

a = -5; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-5)·5
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{26}}{2*-5}=\frac{-2-2\sqrt{26}}{-10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{26}}{2*-5}=\frac{-2+2\sqrt{26}}{-10}
$