-16t^2-22t+338=0

Solution for -16t^2-22t+338=0 equation:

-16t^2-22t+338=0

a = -16; b = -22; c = +338;
Δ = b2-4ac
Δ = -222-4·(-16)·338
Δ = 22116
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:

$\sqrt{\Delta}=\sqrt{22116}=\sqrt{4*5529}=\sqrt{4}*\sqrt{5529}=2\sqrt{5529}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{5529}}{2*-16}=\frac{22-2\sqrt{5529}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{5529}}{2*-16}=\frac{22+2\sqrt{5529}}{-32}
$