p^2+20p+28=9

Solution for p^2+20p+28=9 equation:

p^2+20p+28=9

We move all terms to the left:

p^2+20p+28-(9)=0

We add all the numbers together, and all the variables

p^2+20p+19=0

a = 1; b = 20; c = +19;
Δ = b2-4ac
Δ = 202-4·1·19
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-18}{2*1}=\frac{-38}{2}
=-19
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+18}{2*1}=\frac{-2}{2}
=-1
$