0=4+27t-18t^2

Solution for 0=4+27t-18t^2 equation:

0=4+27t-18t^2

We move all terms to the left:

0-(4+27t-18t^2)=0

We add all the numbers together, and all the variables

-(4+27t-18t^2)=0

We get rid of parentheses

18t^2-27t-4=0

a = 18; b = -27; c = -4;
Δ = b2-4ac
Δ = -272-4·18·(-4)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{113}}{2*18}=\frac{27-3\sqrt{113}}{36}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{113}}{2*18}=\frac{27+3\sqrt{113}}{36}
$