16t2-40t+22=0

Solution for 16t2-40t+22=0 equation:

16t^2-40t+22=0

a = 16; b = -40; c = +22;
Δ = b2-4ac
Δ = -402-4·16·22
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{3}}{2*16}=\frac{40-8\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{3}}{2*16}=\frac{40+8\sqrt{3}}{32}
$