Q=p^2+4p

Solution for Q=p^2+4p equation:

=Q^2+4Q

We move all terms to the left:

-(Q^2+4Q)=0

We get rid of parentheses

-Q^2-4Q=0

We add all the numbers together, and all the variables

-1Q^2-4Q=0

a = -1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-1)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-1}=\frac{0}{-2}
=0
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-1}=\frac{8}{-2}
=-4
$