0=-16t^2+198t-242

Solution for 0=-16t^2+198t-242 equation:

0=-16t^2+198t-242

We move all terms to the left:

0-(-16t^2+198t-242)=0

We add all the numbers together, and all the variables

-(-16t^2+198t-242)=0

We get rid of parentheses

16t^2-198t+242=0

a = 16; b = -198; c = +242;
Δ = b2-4ac
Δ = -1982-4·16·242
Δ = 23716
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23716}=154$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-198)-154}{2*16}=\frac{44}{32}
=1+3/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-198)+154}{2*16}=\frac{352}{32}
=11
$