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

Solution for (2n-1)(4n+2)=0 equation:

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

We multiply parentheses ..

(+8n^2+4n-4n-2)=0

We get rid of parentheses

8n^2+4n-4n-2=0

We add all the numbers together, and all the variables

8n^2-2=0

a = 8; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·8·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*8}=\frac{-8}{16}
=-1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*8}=\frac{8}{16}
=1/2
$