n=4n2-3

Solution for n=4n2-3 equation:

n=4n^2-3

We move all terms to the left:

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

We get rid of parentheses

-4n^2+n+3=0

a = -4; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-4)·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-4}=\frac{-8}{-8}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-4}=\frac{6}{-8}
=-3/4
$