3(p-2)=p(p+2)-6

Solution for 3(p-2)=p(p+2)-6 equation:

3(p-2)=p(p+2)-6

We move all terms to the left:

3(p-2)-(p(p+2)-6)=0

We multiply parentheses

3p-(p(p+2)-6)-6=0


We calculate terms in parentheses: -(p(p+2)-6), so:

p(p+2)-6

We multiply parentheses

p^2+2p-6

Back to the equation:

-(p^2+2p-6)


We get rid of parentheses

-p^2+3p-2p+6-6=0

We add all the numbers together, and all the variables

-1p^2+p=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-1}=\frac{-2}{-2}
=1
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-1}=\frac{0}{-2}
=0
$