0=-16t^2+145t+24

Solution for 0=-16t^2+145t+24 equation:

0=-16t^2+145t+24

We move all terms to the left:

0-(-16t^2+145t+24)=0

We add all the numbers together, and all the variables

-(-16t^2+145t+24)=0

We get rid of parentheses

16t^2-145t-24=0

a = 16; b = -145; c = -24;
Δ = b2-4ac
Δ = -1452-4·16·(-24)
Δ = 22561
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{-(-145)-\sqrt{22561}}{2*16}=\frac{145-\sqrt{22561}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-145)+\sqrt{22561}}{2*16}=\frac{145+\sqrt{22561}}{32}
$