-16t^2+22t+50=0

Solution for -16t^2+22t+50=0 equation:

-16t^2+22t+50=0

a = -16; b = 22; c = +50;
Δ = b2-4ac
Δ = 222-4·(-16)·50
Δ = 3684
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:

$\sqrt{\Delta}=\sqrt{3684}=\sqrt{4*921}=\sqrt{4}*\sqrt{921}=2\sqrt{921}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{921}}{2*-16}=\frac{-22-2\sqrt{921}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{921}}{2*-16}=\frac{-22+2\sqrt{921}}{-32}
$