n(1+n)+3=45

Solution for n(1+n)+3=45 equation:

n(1+n)+3=45

We move all terms to the left:

n(1+n)+3-(45)=0

We add all the numbers together, and all the variables

n(n+1)+3-45=0

We add all the numbers together, and all the variables

n(n+1)-42=0

We multiply parentheses

n^2+n-42=0

a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2}
=6
$