0=-25p^2+1200

Solution for 0=-25p^2+1200 equation:

0=-25p^2+1200

We move all terms to the left:

0-(-25p^2+1200)=0

We add all the numbers together, and all the variables

-(-25p^2+1200)=0

We get rid of parentheses

25p^2-1200=0

a = 25; b = 0; c = -1200;
Δ = b2-4ac
Δ = 02-4·25·(-1200)
Δ = 120000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120000}=\sqrt{40000*3}=\sqrt{40000}*\sqrt{3}=200\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-200\sqrt{3}}{2*25}=\frac{0-200\sqrt{3}}{50}
=-\frac{200\sqrt{3}}{50}
=-4\sqrt{3}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+200\sqrt{3}}{2*25}=\frac{0+200\sqrt{3}}{50}
=\frac{200\sqrt{3}}{50}
=4\sqrt{3}
$