F(p)=(1440-12p)(p-20)

Solution for F(p)=(1440-12p)(p-20) equation:

(F)=(1440-12F)(F-20)

We move all terms to the left:

(F)-((1440-12F)(F-20))=0

We add all the numbers together, and all the variables

F-((-12F+1440)(F-20))=0

We multiply parentheses ..

-((-12F^2+240F+1440F-28800))+F=0


We calculate terms in parentheses: -((-12F^2+240F+1440F-28800)), so:

(-12F^2+240F+1440F-28800)

We get rid of parentheses

-12F^2+240F+1440F-28800

We add all the numbers together, and all the variables

-12F^2+1680F-28800

Back to the equation:

-(-12F^2+1680F-28800)


We get rid of parentheses

12F^2-1680F+F+28800=0

We add all the numbers together, and all the variables

12F^2-1679F+28800=0

a = 12; b = -1679; c = +28800;
Δ = b2-4ac
Δ = -16792-4·12·28800
Δ = 1436641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1679)-\sqrt{1436641}}{2*12}=\frac{1679-\sqrt{1436641}}{24}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1679)+\sqrt{1436641}}{2*12}=\frac{1679+\sqrt{1436641}}{24}
$