-200p^2+300p^2-80=0

Solution for -200p^2+300p^2-80=0 equation:

-200p^2+300p^2-80=0

We add all the numbers together, and all the variables

100p^2-80=0

a = 100; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·100·(-80)
Δ = 32000
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:

$\sqrt{\Delta}=\sqrt{32000}=\sqrt{6400*5}=\sqrt{6400}*\sqrt{5}=80\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{5}}{2*100}=\frac{0-80\sqrt{5}}{200}
=-\frac{80\sqrt{5}}{200}
=-\frac{2\sqrt{5}}{5}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{5}}{2*100}=\frac{0+80\sqrt{5}}{200}
=\frac{80\sqrt{5}}{200}
=\frac{2\sqrt{5}}{5}
$