(p+2)(7p+2)=0

Solution for (p+2)(7p+2)=0 equation:

(p+2)(7p+2)=0

We multiply parentheses ..

(+7p^2+2p+14p+4)=0

We get rid of parentheses

7p^2+2p+14p+4=0

We add all the numbers together, and all the variables

7p^2+16p+4=0

a = 7; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·7·4
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{-(16)-12}{2*7}=\frac{-28}{14}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12}{2*7}=\frac{-4}{14}
=-2/7
$