p^2+20p+102=6

Solution for p^2+20p+102=6 equation:

p^2+20p+102=6

We move all terms to the left:

p^2+20p+102-(6)=0

We add all the numbers together, and all the variables

p^2+20p+96=0

a = 1; b = 20; c = +96;
Δ = b2-4ac
Δ = 202-4·1·96
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{-(20)-4}{2*1}=\frac{-24}{2}
=-12
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*1}=\frac{-16}{2}
=-8
$