-16t^2+54.4897615t+5=0

Solution for -16t^2+54.4897615t+5=0 equation:

-16t^2+54.4897615t+5=0

a = -16; b = 54.4897615; c = +5;
Δ = b2-4ac
Δ = 54.48976152-4·(-16)·5
Δ = 3289.1341083269
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{-(54.4897615)-\sqrt{3289.1341083269}}{2*-16}=\frac{-54.4897615-\sqrt{3289.1341083269}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54.4897615)+\sqrt{3289.1341083269}}{2*-16}=\frac{-54.4897615+\sqrt{3289.1341083269}}{-32}
$