(4n-1)(n+1)=0

Solution for (4n-1)(n+1)=0 equation:

(4n-1)(n+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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