n=4(n/2)^2

Solution for n=4(n/2)^2 equation:

n=4(n/2)^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

2n^2+n=0

a = 2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*2}=\frac{-2}{4}
=-1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*2}=\frac{0}{4}
=0
$