45=n*(n+4)

Solution for 45=n*(n+4) equation:

45=n(n+4)

We move all terms to the left:

45-(n(n+4))=0


We calculate terms in parentheses: -(n(n+4)), so:

n(n+4)

We multiply parentheses

n^2+4n

Back to the equation:

-(n^2+4n)


We get rid of parentheses

-n^2-4n+45=0

We add all the numbers together, and all the variables

-1n^2-4n+45=0

a = -1; b = -4; c = +45;
Δ = b2-4ac
Δ = -42-4·(-1)·45
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*-1}=\frac{-10}{-2}
=+5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*-1}=\frac{18}{-2}
=-9
$