0=-16t^2+155t+41

Solution for 0=-16t^2+155t+41 equation:

0=-16t^2+155t+41

We move all terms to the left:

0-(-16t^2+155t+41)=0

We add all the numbers together, and all the variables

-(-16t^2+155t+41)=0

We get rid of parentheses

16t^2-155t-41=0

a = 16; b = -155; c = -41;
Δ = b2-4ac
Δ = -1552-4·16·(-41)
Δ = 26649
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:

$\sqrt{\Delta}=\sqrt{26649}=\sqrt{81*329}=\sqrt{81}*\sqrt{329}=9\sqrt{329}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-155)-9\sqrt{329}}{2*16}=\frac{155-9\sqrt{329}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-155)+9\sqrt{329}}{2*16}=\frac{155+9\sqrt{329}}{32}
$