0=-16t^2+304

Solution for 0=-16t^2+304 equation:

0=-16t^2+304

We move all terms to the left:

0-(-16t^2+304)=0

We add all the numbers together, and all the variables

-(-16t^2+304)=0

We get rid of parentheses

16t^2-304=0

a = 16; b = 0; c = -304;
Δ = b2-4ac
Δ = 02-4·16·(-304)
Δ = 19456
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:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{19}}{2*16}=\frac{0-32\sqrt{19}}{32}
=-\frac{32\sqrt{19}}{32}
=-\sqrt{19}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{19}}{2*16}=\frac{0+32\sqrt{19}}{32}
=\frac{32\sqrt{19}}{32}
=\sqrt{19}
$