t=3+18t-18t^2

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

t=3+18t-18t^2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

18t^2-17t-3=0

a = 18; b = -17; c = -3;
Δ = b2-4ac
Δ = -172-4·18·(-3)
Δ = 505
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{505}}{2*18}=\frac{17-\sqrt{505}}{36}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{505}}{2*18}=\frac{17+\sqrt{505}}{36}
$