3n+4=27n2

Solution for 3n+4=27n2 equation:

3n+4=27n^2

We move all terms to the left:

3n+4-(27n^2)=0

determiningTheFunctionDomain
-27n^2+3n+4=0

a = -27; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-27)·4
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-27}=\frac{-24}{-54}
=4/9
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-27}=\frac{18}{-54}
=-1/3
$