16t^2-128t+211=0

Solution for 16t^2-128t+211=0 equation:

16t^2-128t+211=0

a = 16; b = -128; c = +211;
Δ = b2-4ac
Δ = -1282-4·16·211
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-24\sqrt{5}}{2*16}=\frac{128-24\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+24\sqrt{5}}{2*16}=\frac{128+24\sqrt{5}}{32}
$