18=2+36t-16t^2

Solution for 18=2+36t-16t^2 equation:

18=2+36t-16t^2

We move all terms to the left:

18-(2+36t-16t^2)=0

We get rid of parentheses

16t^2-36t-2+18=0

We add all the numbers together, and all the variables

16t^2-36t+16=0

a = 16; b = -36; c = +16;
Δ = b2-4ac
Δ = -362-4·16·16
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{17}}{2*16}=\frac{36-4\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{17}}{2*16}=\frac{36+4\sqrt{17}}{32}
$