4(n)=-2n2+4

Solution for 4(n)=-2n2+4 equation:

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

We move all terms to the left:

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

We get rid of parentheses

2n^2+4n-4=0

a = 2; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·2·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*2}=\frac{-4-4\sqrt{3}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*2}=\frac{-4+4\sqrt{3}}{4}
$