-16t^2+22t-5=0

Solution for -16t^2+22t-5=0 equation:

-16t^2+22t-5=0

a = -16; b = 22; c = -5;
Δ = b2-4ac
Δ = 222-4·(-16)·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{41}}{2*-16}=\frac{-22-2\sqrt{41}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{41}}{2*-16}=\frac{-22+2\sqrt{41}}{-32}
$