n(3^-5)=3

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

n(3^-5)=3

We move all terms to the left:

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

We multiply parentheses

3n^2-5n-3=0

a = 3; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·3·(-3)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{61}}{2*3}=\frac{5-\sqrt{61}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{61}}{2*3}=\frac{5+\sqrt{61}}{6}
$