0=-16t^2+153t+98

Solution for 0=-16t^2+153t+98 equation:

0=-16t^2+153t+98

We move all terms to the left:

0-(-16t^2+153t+98)=0

We add all the numbers together, and all the variables

-(-16t^2+153t+98)=0

We get rid of parentheses

16t^2-153t-98=0

a = 16; b = -153; c = -98;
Δ = b2-4ac
Δ = -1532-4·16·(-98)
Δ = 29681
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{-(-153)-\sqrt{29681}}{2*16}=\frac{153-\sqrt{29681}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-153)+\sqrt{29681}}{2*16}=\frac{153+\sqrt{29681}}{32}
$