2n+3=n^2

Solution for 2n+3=n^2 equation:

2n+3=n^2

We move all terms to the left:

2n+3-(n^2)=0

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

We add all the numbers together, and all the variables

-1n^2+2n+3=0

a = -1; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*-1}=\frac{-6}{-2}
=+3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-1}=\frac{2}{-2}
=-1
$