-16t^2+3t+0.8=0

Solution for -16t^2+3t+0.8=0 equation:

-16t^2+3t+0.8=0

a = -16; b = 3; c = +0.8;
Δ = b2-4ac
Δ = 32-4·(-16)·0.8
Δ = 60.2
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{-(3)-\sqrt{60.2}}{2*-16}=\frac{-3-\sqrt{60.2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{60.2}}{2*-16}=\frac{-3+\sqrt{60.2}}{-32}
$