0=-16t^2+29t-11

Solution for 0=-16t^2+29t-11 equation:

0=-16t^2+29t-11

We move all terms to the left:

0-(-16t^2+29t-11)=0

We add all the numbers together, and all the variables

-(-16t^2+29t-11)=0

We get rid of parentheses

16t^2-29t+11=0

a = 16; b = -29; c = +11;
Δ = b2-4ac
Δ = -292-4·16·11
Δ = 137
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{-(-29)-\sqrt{137}}{2*16}=\frac{29-\sqrt{137}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{137}}{2*16}=\frac{29+\sqrt{137}}{32}
$