-p^2=-16p+28

Solution for -p^2=-16p+28 equation:

-p^2=-16p+28

We move all terms to the left:

-p^2-(-16p+28)=0

We add all the numbers together, and all the variables

-1p^2-(-16p+28)=0

We get rid of parentheses

-1p^2+16p-28=0

a = -1; b = 16; c = -28;
Δ = b2-4ac
Δ = 162-4·(-1)·(-28)
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*-1}=\frac{-28}{-2}
=+14
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12}{2*-1}=\frac{-4}{-2}
=+2
$