9n^2+3=67

Solution for 9n^2+3=67 equation:

9n^2+3=67

We move all terms to the left:

9n^2+3-(67)=0

We add all the numbers together, and all the variables

9n^2-64=0

a = 9; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·9·(-64)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*9}=\frac{-48}{18}
=-2+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*9}=\frac{48}{18}
=2+2/3
$