-16t^2+43.3t+6=0

Solution for -16t^2+43.3t+6=0 equation:

-16t^2+43.3t+6=0

a = -16; b = 43.3; c = +6;
Δ = b2-4ac
Δ = 43.32-4·(-16)·6
Δ = 2258.89
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{-(43.3)-\sqrt{2258.89}}{2*-16}=\frac{-43.3-\sqrt{2258.89}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43.3)+\sqrt{2258.89}}{2*-16}=\frac{-43.3+\sqrt{2258.89}}{-32}
$