4(n)=n2+2n=5

Solution for 4(n)=n2+2n=5 equation:

4(n)=n2+2n=5

We move all terms to the left:

4(n)-(n2+2n)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1n^2+2n=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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