0=16t^2-96t+4

Solution for 0=16t^2-96t+4 equation:

0=16t^2-96t+4

We move all terms to the left:

0-(16t^2-96t+4)=0

We add all the numbers together, and all the variables

-(16t^2-96t+4)=0

We get rid of parentheses

-16t^2+96t-4=0

a = -16; b = 96; c = -4;
Δ = b2-4ac
Δ = 962-4·(-16)·(-4)
Δ = 8960
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:

$\sqrt{\Delta}=\sqrt{8960}=\sqrt{256*35}=\sqrt{256}*\sqrt{35}=16\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{35}}{2*-16}=\frac{-96-16\sqrt{35}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{35}}{2*-16}=\frac{-96+16\sqrt{35}}{-32}
$