0=3t^2-18t-16

Solution for 0=3t^2-18t-16 equation:

0=3t^2-18t-16

We move all terms to the left:

0-(3t^2-18t-16)=0

We add all the numbers together, and all the variables

-(3t^2-18t-16)=0

We get rid of parentheses

-3t^2+18t+16=0

a = -3; b = 18; c = +16;
Δ = b2-4ac
Δ = 182-4·(-3)·16
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{129}}{2*-3}=\frac{-18-2\sqrt{129}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{129}}{2*-3}=\frac{-18+2\sqrt{129}}{-6}
$