P=-1028t^2+90000t+50000

Solution for P=-1028t^2+90000t+50000 equation:

=-1028P^2+90000P+50000

We move all terms to the left:

-(-1028P^2+90000P+50000)=0

We get rid of parentheses

1028P^2-90000P-50000=0

a = 1028; b = -90000; c = -50000;
Δ = b2-4ac
Δ = -900002-4·1028·(-50000)
Δ = 8305600000
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:

$\sqrt{\Delta}=\sqrt{8305600000}=\sqrt{160000*51910}=\sqrt{160000}*\sqrt{51910}=400\sqrt{51910}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90000)-400\sqrt{51910}}{2*1028}=\frac{90000-400\sqrt{51910}}{2056}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90000)+400\sqrt{51910}}{2*1028}=\frac{90000+400\sqrt{51910}}{2056}
$