2(p+3)(p+4)=4

Solution for 2(p+3)(p+4)=4 equation:

2(p+3)(p+4)=4

We move all terms to the left:

2(p+3)(p+4)-(4)=0

We multiply parentheses ..

2(+p^2+4p+3p+12)-4=0

We multiply parentheses

2p^2+8p+6p+24-4=0

We add all the numbers together, and all the variables

2p^2+14p+20=0

a = 2; b = 14; c = +20;
Δ = b2-4ac
Δ = 142-4·2·20
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*2}=\frac{-20}{4}
=-5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*2}=\frac{-8}{4}
=-2
$