-16t^2+10.5t+5=0

Solution for -16t^2+10.5t+5=0 equation:

-16t^2+10.5t+5=0

a = -16; b = 10.5; c = +5;
Δ = b2-4ac
Δ = 10.52-4·(-16)·5
Δ = 430.25
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{-(10.5)-\sqrt{430.25}}{2*-16}=\frac{-10.5-\sqrt{430.25}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10.5)+\sqrt{430.25}}{2*-16}=\frac{-10.5+\sqrt{430.25}}{-32}
$