0=-16t^2+105t-89

Solution for 0=-16t^2+105t-89 equation:

0=-16t^2+105t-89

We move all terms to the left:

0-(-16t^2+105t-89)=0

We add all the numbers together, and all the variables

-(-16t^2+105t-89)=0

We get rid of parentheses

16t^2-105t+89=0

a = 16; b = -105; c = +89;
Δ = b2-4ac
Δ = -1052-4·16·89
Δ = 5329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5329}=73$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-105)-73}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+73}{2*16}=\frac{178}{32}
=5+9/16
$