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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-121t-10=0

a = 16; b = -121; c = -10;
Δ = b2-4ac
Δ = -1212-4·16·(-10)
Δ = 15281
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{-(-121)-\sqrt{15281}}{2*16}=\frac{121-\sqrt{15281}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-121)+\sqrt{15281}}{2*16}=\frac{121+\sqrt{15281}}{32}
$