0=-16t^2+16t+38

Solution for 0=-16t^2+16t+38 equation:

0=-16t^2+16t+38

We move all terms to the left:

0-(-16t^2+16t+38)=0

We add all the numbers together, and all the variables

-(-16t^2+16t+38)=0

We get rid of parentheses

16t^2-16t-38=0

a = 16; b = -16; c = -38;
Δ = b2-4ac
Δ = -162-4·16·(-38)
Δ = 2688
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:

$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{42}}{2*16}=\frac{16-8\sqrt{42}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{42}}{2*16}=\frac{16+8\sqrt{42}}{32}
$