P(0)=-16t^2+290

Solution for P(0)=-16t^2+290 equation:

(0)=-16P^2+290

We move all terms to the left:

(0)-(-16P^2+290)=0

We add all the numbers together, and all the variables

-(-16P^2+290)=0

We get rid of parentheses

16P^2-290=0

a = 16; b = 0; c = -290;
Δ = b2-4ac
Δ = 02-4·16·(-290)
Δ = 18560
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:

$\sqrt{\Delta}=\sqrt{18560}=\sqrt{64*290}=\sqrt{64}*\sqrt{290}=8\sqrt{290}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{290}}{2*16}=\frac{0-8\sqrt{290}}{32}
=-\frac{8\sqrt{290}}{32}
=-\frac{\sqrt{290}}{4}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{290}}{2*16}=\frac{0+8\sqrt{290}}{32}
=\frac{8\sqrt{290}}{32}
=\frac{\sqrt{290}}{4}
$