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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

3t^2-18t-2=0

a = 3; b = -18; c = -2;
Δ = b2-4ac
Δ = -182-4·3·(-2)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*3}=\frac{18-2\sqrt{87}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*3}=\frac{18+2\sqrt{87}}{6}
$