-16t^2+160t+300=500

Solution for -16t^2+160t+300=500 equation:

-16t^2+160t+300=500

We move all terms to the left:

-16t^2+160t+300-(500)=0

We add all the numbers together, and all the variables

-16t^2+160t-200=0

a = -16; b = 160; c = -200;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-200)
Δ = 12800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-80\sqrt{2}}{2*-16}=\frac{-160-80\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+80\sqrt{2}}{2*-16}=\frac{-160+80\sqrt{2}}{-32}
$