4n^2=27

Solution for 4n^2=27 equation:

4n^2=27

We move all terms to the left:

4n^2-(27)=0

a = 4; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·4·(-27)
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*4}=\frac{0-12\sqrt{3}}{8}
=-\frac{12\sqrt{3}}{8}
=-\frac{3\sqrt{3}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*4}=\frac{0+12\sqrt{3}}{8}
=\frac{12\sqrt{3}}{8}
=\frac{3\sqrt{3}}{2}
$