6=16t2+12t+4

Solution for 6=16t2+12t+4 equation:

6=16t^2+12t+4

We move all terms to the left:

6-(16t^2+12t+4)=0

We get rid of parentheses

-16t^2-12t-4+6=0

We add all the numbers together, and all the variables

-16t^2-12t+2=0

a = -16; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·(-16)·2
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{17}}{2*-16}=\frac{12-4\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{17}}{2*-16}=\frac{12+4\sqrt{17}}{-32}
$