0=-16t^2+64t+312

Solution for 0=-16t^2+64t+312 equation:

0=-16t^2+64t+312

We move all terms to the left:

0-(-16t^2+64t+312)=0

We add all the numbers together, and all the variables

-(-16t^2+64t+312)=0

We get rid of parentheses

16t^2-64t-312=0

a = 16; b = -64; c = -312;
Δ = b2-4ac
Δ = -642-4·16·(-312)
Δ = 24064
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:

$\sqrt{\Delta}=\sqrt{24064}=\sqrt{256*94}=\sqrt{256}*\sqrt{94}=16\sqrt{94}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-16\sqrt{94}}{2*16}=\frac{64-16\sqrt{94}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+16\sqrt{94}}{2*16}=\frac{64+16\sqrt{94}}{32}
$