(5n)^2=n^2+3^2=n^2+9

Solution for (5n)^2=n^2+3^2=n^2+9 equation:

(5n)^2=n^2+3^2=n^2+9

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4n^2-9=0

a = 4; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·4·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*4}=\frac{-12}{8}
=-1+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*4}=\frac{12}{8}
=1+1/2
$