0=-16t^2+38t+4

Solution for 0=-16t^2+38t+4 equation:

0=-16t^2+38t+4

We move all terms to the left:

0-(-16t^2+38t+4)=0

We add all the numbers together, and all the variables

-(-16t^2+38t+4)=0

We get rid of parentheses

16t^2-38t-4=0

a = 16; b = -38; c = -4;
Δ = b2-4ac
Δ = -382-4·16·(-4)
Δ = 1700
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:

$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-10\sqrt{17}}{2*16}=\frac{38-10\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+10\sqrt{17}}{2*16}=\frac{38+10\sqrt{17}}{32}
$