12-28p=5p^2

Solution for 12-28p=5p^2 equation:

12-28p=5p^2

We move all terms to the left:

12-28p-(5p^2)=0

determiningTheFunctionDomain
-5p^2-28p+12=0

a = -5; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·(-5)·12
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-32}{2*-5}=\frac{-4}{-10}
=2/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+32}{2*-5}=\frac{60}{-10}
=-6
$