10=-16t^2+29t+3

Solution for 10=-16t^2+29t+3 equation:

10=-16t^2+29t+3

We move all terms to the left:

10-(-16t^2+29t+3)=0

We get rid of parentheses

16t^2-29t-3+10=0

We add all the numbers together, and all the variables

16t^2-29t+7=0

a = 16; b = -29; c = +7;
Δ = b2-4ac
Δ = -292-4·16·7
Δ = 393
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{393}}{2*16}=\frac{29-\sqrt{393}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{393}}{2*16}=\frac{29+\sqrt{393}}{32}
$