4n-n^2=2

Solution for 4n-n^2=2 equation:

4n-n^2=2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

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

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