0=-16t^2+122t+10

Solution for 0=-16t^2+122t+10 equation:

0=-16t^2+122t+10

We move all terms to the left:

0-(-16t^2+122t+10)=0

We add all the numbers together, and all the variables

-(-16t^2+122t+10)=0

We get rid of parentheses

16t^2-122t-10=0

a = 16; b = -122; c = -10;
Δ = b2-4ac
Δ = -1222-4·16·(-10)
Δ = 15524
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:

$\sqrt{\Delta}=\sqrt{15524}=\sqrt{4*3881}=\sqrt{4}*\sqrt{3881}=2\sqrt{3881}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-122)-2\sqrt{3881}}{2*16}=\frac{122-2\sqrt{3881}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-122)+2\sqrt{3881}}{2*16}=\frac{122+2\sqrt{3881}}{32}
$