0=3t^2-16t+4

Solution for 0=3t^2-16t+4 equation:

0=3t^2-16t+4

We move all terms to the left:

0-(3t^2-16t+4)=0

We add all the numbers together, and all the variables

-(3t^2-16t+4)=0

We get rid of parentheses

-3t^2+16t-4=0

a = -3; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·(-3)·(-4)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{13}}{2*-3}=\frac{-16-4\sqrt{13}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{13}}{2*-3}=\frac{-16+4\sqrt{13}}{-6}
$