-16t^2+209t+9=0

Solution for -16t^2+209t+9=0 equation:

-16t^2+209t+9=0

a = -16; b = 209; c = +9;
Δ = b2-4ac
Δ = 2092-4·(-16)·9
Δ = 44257
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{-(209)-\sqrt{44257}}{2*-16}=\frac{-209-\sqrt{44257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(209)+\sqrt{44257}}{2*-16}=\frac{-209+\sqrt{44257}}{-32}
$