-16t^2-29t+491=122

Solution for -16t^2-29t+491=122 equation:

-16t^2-29t+491=122

We move all terms to the left:

-16t^2-29t+491-(122)=0

We add all the numbers together, and all the variables

-16t^2-29t+369=0

a = -16; b = -29; c = +369;
Δ = b2-4ac
Δ = -292-4·(-16)·369
Δ = 24457
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{24457}}{2*-16}=\frac{29-\sqrt{24457}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{24457}}{2*-16}=\frac{29+\sqrt{24457}}{-32}
$