-16t^2+62t+29=0

Solution for -16t^2+62t+29=0 equation:

-16t^2+62t+29=0

a = -16; b = 62; c = +29;
Δ = b2-4ac
Δ = 622-4·(-16)·29
Δ = 5700
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:

$\sqrt{\Delta}=\sqrt{5700}=\sqrt{100*57}=\sqrt{100}*\sqrt{57}=10\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-10\sqrt{57}}{2*-16}=\frac{-62-10\sqrt{57}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+10\sqrt{57}}{2*-16}=\frac{-62+10\sqrt{57}}{-32}
$