5=(p+5)(p+4)

Solution for 5=(p+5)(p+4) equation:

5=(p+5)(p+4)

We move all terms to the left:

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

We multiply parentheses ..

-((+p^2+4p+5p+20))+5=0


We calculate terms in parentheses: -((+p^2+4p+5p+20)), so:

(+p^2+4p+5p+20)

We get rid of parentheses

p^2+4p+5p+20

We add all the numbers together, and all the variables

p^2+9p+20

Back to the equation:

-(p^2+9p+20)


We get rid of parentheses

-p^2-9p-20+5=0

We add all the numbers together, and all the variables

-1p^2-9p-15=0

a = -1; b = -9; c = -15;
Δ = b2-4ac
Δ = -92-4·(-1)·(-15)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{21}}{2*-1}=\frac{9-\sqrt{21}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{21}}{2*-1}=\frac{9+\sqrt{21}}{-2}
$