132=-16t^2+132t+18

Solution for 132=-16t^2+132t+18 equation:

132=-16t^2+132t+18

We move all terms to the left:

132-(-16t^2+132t+18)=0

We get rid of parentheses

16t^2-132t-18+132=0

We add all the numbers together, and all the variables

16t^2-132t+114=0

a = 16; b = -132; c = +114;
Δ = b2-4ac
Δ = -1322-4·16·114
Δ = 10128
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:

$\sqrt{\Delta}=\sqrt{10128}=\sqrt{16*633}=\sqrt{16}*\sqrt{633}=4\sqrt{633}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-132)-4\sqrt{633}}{2*16}=\frac{132-4\sqrt{633}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-132)+4\sqrt{633}}{2*16}=\frac{132+4\sqrt{633}}{32}
$