0=197-21t-16t^2

Solution for 0=197-21t-16t^2 equation:

0=197-21t-16t^2

We move all terms to the left:

0-(197-21t-16t^2)=0

We add all the numbers together, and all the variables

-(197-21t-16t^2)=0

We get rid of parentheses

16t^2+21t-197=0

a = 16; b = 21; c = -197;
Δ = b2-4ac
Δ = 212-4·16·(-197)
Δ = 13049
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{-(21)-\sqrt{13049}}{2*16}=\frac{-21-\sqrt{13049}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{13049}}{2*16}=\frac{-21+\sqrt{13049}}{32}
$