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

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

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

We use the square of the difference formula

16p^2-1=0

a = 16; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·16·(-1)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*16}=\frac{-8}{32}
=-1/4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*16}=\frac{8}{32}
=1/4
$