-16t^2+55t+190=0

Solution for -16t^2+55t+190=0 equation:

-16t^2+55t+190=0

a = -16; b = 55; c = +190;
Δ = b2-4ac
Δ = 552-4·(-16)·190
Δ = 15185
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{-(55)-\sqrt{15185}}{2*-16}=\frac{-55-\sqrt{15185}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+\sqrt{15185}}{2*-16}=\frac{-55+\sqrt{15185}}{-32}
$