n/4=3n^2-1

Solution for n/4=3n^2-1 equation:

n/4=3n^2-1

We move all terms to the left:

n/4-(3n^2-1)=0

We get rid of parentheses

-3n^2+n/4+1=0

We multiply all the terms by the denominator


-3n^2*4+n+1*4=0

We add all the numbers together, and all the variables

-3n^2*4+n+4=0

Wy multiply elements

-12n^2+n+4=0

a = -12; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·(-12)·4
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{193}}{2*-12}=\frac{-1-\sqrt{193}}{-24}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*-12}=\frac{-1+\sqrt{193}}{-24}
$