2=-16t^2+129t

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

2=-16t^2+129t

We move all terms to the left:

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

We get rid of parentheses

16t^2-129t+2=0

a = 16; b = -129; c = +2;
Δ = b2-4ac
Δ = -1292-4·16·2
Δ = 16513
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:

$\sqrt{\Delta}=\sqrt{16513}=\sqrt{49*337}=\sqrt{49}*\sqrt{337}=7\sqrt{337}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-129)-7\sqrt{337}}{2*16}=\frac{129-7\sqrt{337}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-129)+7\sqrt{337}}{2*16}=\frac{129+7\sqrt{337}}{32}
$