(2p^2+4p)/(p^2+4p+3)=0

Solution for (2p^2+4p)/(p^2+4p+3)=0 equation:

(2p^2+4p)/(p^2+4p+3)=0


Domain of the equation: (p^2+4p+3)!=0

We move all terms containing p to the left, all other terms to the right

p^2+4p!=-3

p∈R


We multiply all the terms by the denominator


(2p^2+4p)=0

We get rid of parentheses

2p^2+4p=0

a = 2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·2·0
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{-(4)-4}{2*2}=\frac{-8}{4}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*2}=\frac{0}{4}
=0
$