16t^2-96t+16=0

Solution for 16t^2-96t+16=0 equation:

16t^2-96t+16=0

a = 16; b = -96; c = +16;
Δ = b2-4ac
Δ = -962-4·16·16
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-64\sqrt{2}}{2*16}=\frac{96-64\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+64\sqrt{2}}{2*16}=\frac{96+64\sqrt{2}}{32}
$