237=-16t^2+129t+18

Solution for 237=-16t^2+129t+18 equation:

237=-16t^2+129t+18

We move all terms to the left:

237-(-16t^2+129t+18)=0

We get rid of parentheses

16t^2-129t-18+237=0

We add all the numbers together, and all the variables

16t^2-129t+219=0

a = 16; b = -129; c = +219;
Δ = b2-4ac
Δ = -1292-4·16·219
Δ = 2625
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:

$\sqrt{\Delta}=\sqrt{2625}=\sqrt{25*105}=\sqrt{25}*\sqrt{105}=5\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-129)-5\sqrt{105}}{2*16}=\frac{129-5\sqrt{105}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-129)+5\sqrt{105}}{2*16}=\frac{129+5\sqrt{105}}{32}
$