-16t2+320t-1538=0

Solution for -16t2+320t-1538=0 equation:

-16t^2+320t-1538=0

a = -16; b = 320; c = -1538;
Δ = b2-4ac
Δ = 3202-4·(-16)·(-1538)
Δ = 3968
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3968}=\sqrt{64*62}=\sqrt{64}*\sqrt{62}=8\sqrt{62}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-8\sqrt{62}}{2*-16}=\frac{-320-8\sqrt{62}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+8\sqrt{62}}{2*-16}=\frac{-320+8\sqrt{62}}{-32}
$