-16t^2+62.7974705t=0

Solution for -16t^2+62.7974705t=0 equation:

-16t^2+62.7974705t=0

a = -16; b = 62.7974705; c = 0;
Δ = b2-4ac
Δ = 62.79747052-4·(-16)·0
Δ = 3943.5223011984
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{-(62.7974705)-\sqrt{3943.5223011984}}{2*-16}=\frac{-62.7974705-\sqrt{3943.5223011984}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62.7974705)+\sqrt{3943.5223011984}}{2*-16}=\frac{-62.7974705+\sqrt{3943.5223011984}}{-32}
$