-16t^2+45t+105=0

Solution for -16t^2+45t+105=0 equation:

-16t^2+45t+105=0

a = -16; b = 45; c = +105;
Δ = b2-4ac
Δ = 452-4·(-16)·105
Δ = 8745
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{-(45)-\sqrt{8745}}{2*-16}=\frac{-45-\sqrt{8745}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{8745}}{2*-16}=\frac{-45+\sqrt{8745}}{-32}
$