0=3t^2+18t+4

Solution for 0=3t^2+18t+4 equation:

0=3t^2+18t+4

We move all terms to the left:

0-(3t^2+18t+4)=0

We add all the numbers together, and all the variables

-(3t^2+18t+4)=0

We get rid of parentheses

-3t^2-18t-4=0

a = -3; b = -18; c = -4;
Δ = b2-4ac
Δ = -182-4·(-3)·(-4)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{69}}{2*-3}=\frac{18-2\sqrt{69}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{69}}{2*-3}=\frac{18+2\sqrt{69}}{-6}
$