16t^2-34t+16=0

Solution for 16t^2-34t+16=0 equation:

16t^2-34t+16=0

a = 16; b = -34; c = +16;
Δ = b2-4ac
Δ = -342-4·16·16
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{33}}{2*16}=\frac{34-2\sqrt{33}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{33}}{2*16}=\frac{34+2\sqrt{33}}{32}
$