4(p+1)(1-p)=3

Solution for 4(p+1)(1-p)=3 equation:

4(p+1)(1-p)=3

We move all terms to the left:

4(p+1)(1-p)-(3)=0

We add all the numbers together, and all the variables

4(p+1)(-1p+1)-3=0

We multiply parentheses ..

4(-1p^2+p-1p+1)-3=0

We multiply parentheses

-4p^2+4p-4p+4-3=0

We add all the numbers together, and all the variables

-4p^2+1=0

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

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