9n^2=18n+27

Solution for 9n^2=18n+27 equation:

9n^2=18n+27

We move all terms to the left:

9n^2-(18n+27)=0

We get rid of parentheses

9n^2-18n-27=0

a = 9; b = -18; c = -27;
Δ = b2-4ac
Δ = -182-4·9·(-27)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-36}{2*9}=\frac{-18}{18}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+36}{2*9}=\frac{54}{18}
=3
$