0=-16t^2+128t+50

Solution for 0=-16t^2+128t+50 equation:

0=-16t^2+128t+50

We move all terms to the left:

0-(-16t^2+128t+50)=0

We add all the numbers together, and all the variables

-(-16t^2+128t+50)=0

We get rid of parentheses

16t^2-128t-50=0

a = 16; b = -128; c = -50;
Δ = b2-4ac
Δ = -1282-4·16·(-50)
Δ = 19584
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:

$\sqrt{\Delta}=\sqrt{19584}=\sqrt{576*34}=\sqrt{576}*\sqrt{34}=24\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-24\sqrt{34}}{2*16}=\frac{128-24\sqrt{34}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+24\sqrt{34}}{2*16}=\frac{128+24\sqrt{34}}{32}
$