0=3t^2+16t+12

Solution for 0=3t^2+16t+12 equation:

0=3t^2+16t+12

We move all terms to the left:

0-(3t^2+16t+12)=0

We add all the numbers together, and all the variables

-(3t^2+16t+12)=0

We get rid of parentheses

-3t^2-16t-12=0

a = -3; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·(-3)·(-12)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{7}}{2*-3}=\frac{16-4\sqrt{7}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{7}}{2*-3}=\frac{16+4\sqrt{7}}{-6}
$