-16t^2-29+491=122

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

-16t^2-29+491=122

We move all terms to the left:

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

We add all the numbers together, and all the variables

-16t^2+340=0

a = -16; b = 0; c = +340;
Δ = b2-4ac
Δ = 02-4·(-16)·340
Δ = 21760
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:

$\sqrt{\Delta}=\sqrt{21760}=\sqrt{256*85}=\sqrt{256}*\sqrt{85}=16\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{85}}{2*-16}=\frac{0-16\sqrt{85}}{-32}
=-\frac{16\sqrt{85}}{-32}
=-\frac{\sqrt{85}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{85}}{2*-16}=\frac{0+16\sqrt{85}}{-32}
=\frac{16\sqrt{85}}{-32}
=\frac{\sqrt{85}}{-2}
$