4^2-4(p)(5-p)=0

Solution for 4^2-4(p)(5-p)=0 equation:

4^2-4(p)(5-p)=0

We add all the numbers together, and all the variables

-4p(-1p+5)+4^2=0

We add all the numbers together, and all the variables

-4p(-1p+5)+16=0

We multiply parentheses

4p^2-20p+16=0

a = 4; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·4·16
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*4}=\frac{8}{8}
=1
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*4}=\frac{32}{8}
=4
$