0=16t^2+19t-5

Solution for 0=16t^2+19t-5 equation:

0=16t^2+19t-5

We move all terms to the left:

0-(16t^2+19t-5)=0

We add all the numbers together, and all the variables

-(16t^2+19t-5)=0

We get rid of parentheses

-16t^2-19t+5=0

a = -16; b = -19; c = +5;
Δ = b2-4ac
Δ = -192-4·(-16)·5
Δ = 681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{681}}{2*-16}=\frac{19-\sqrt{681}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{681}}{2*-16}=\frac{19+\sqrt{681}}{-32}
$