0=5+38t+16t2

Solution for 0=5+38t+16t2 equation:

0=5+38t+16t^2

We move all terms to the left:

0-(5+38t+16t^2)=0

We add all the numbers together, and all the variables

-(5+38t+16t^2)=0

We get rid of parentheses

-16t^2-38t-5=0

a = -16; b = -38; c = -5;
Δ = b2-4ac
Δ = -382-4·(-16)·(-5)
Δ = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{281}}{2*-16}=\frac{38-2\sqrt{281}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{281}}{2*-16}=\frac{38+2\sqrt{281}}{-32}
$