5n=n(n+3)-5

Solution for 5n=n(n+3)-5 equation:

5n=n(n+3)-5

We move all terms to the left:

5n-(n(n+3)-5)=0


We calculate terms in parentheses: -(n(n+3)-5), so:

n(n+3)-5

We multiply parentheses

n^2+3n-5

Back to the equation:

-(n^2+3n-5)


We get rid of parentheses

-n^2+5n-3n+5=0

We add all the numbers together, and all the variables

-1n^2+2n+5=0

a = -1; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-1)·5
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*-1}=\frac{-2-2\sqrt{6}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*-1}=\frac{-2+2\sqrt{6}}{-2}
$