4n^2-3n=1

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

4n^2-3n=1

We move all terms to the left:

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

a = 4; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·4·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*4}=\frac{-2}{8}
=-1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*4}=\frac{8}{8}
=1
$