6(n^2+3)=57

Solution for 6(n^2+3)=57 equation:

6(n^2+3)=57

We move all terms to the left:

6(n^2+3)-(57)=0

We multiply parentheses

6n^2+18-57=0

We add all the numbers together, and all the variables

6n^2-39=0

a = 6; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·6·(-39)
Δ = 936
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:

$\sqrt{\Delta}=\sqrt{936}=\sqrt{36*26}=\sqrt{36}*\sqrt{26}=6\sqrt{26}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{26}}{2*6}=\frac{0-6\sqrt{26}}{12}
=-\frac{6\sqrt{26}}{12}
=-\frac{\sqrt{26}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{26}}{2*6}=\frac{0+6\sqrt{26}}{12}
=\frac{6\sqrt{26}}{12}
=\frac{\sqrt{26}}{2}
$