p^2-p-20=(p-5)

Solution for p^2-p-20=(p-5) equation:

p^2-p-20=(p-5)

We move all terms to the left:

p^2-p-20-((p-5))=0

We add all the numbers together, and all the variables

p^2-1p-((p-5))-20=0


We calculate terms in parentheses: -((p-5)), so:

(p-5)

We get rid of parentheses

p-5

Back to the equation:

-(p-5)


We get rid of parentheses

p^2-1p-p+5-20=0

We add all the numbers together, and all the variables

p^2-2p-15=0

a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
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{-(-2)-8}{2*1}=\frac{-6}{2}
=-3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2}
=5
$