0=-16t^2+40t+54

Solution for 0=-16t^2+40t+54 equation:

0=-16t^2+40t+54

We move all terms to the left:

0-(-16t^2+40t+54)=0

We add all the numbers together, and all the variables

-(-16t^2+40t+54)=0

We get rid of parentheses

16t^2-40t-54=0

a = 16; b = -40; c = -54;
Δ = b2-4ac
Δ = -402-4·16·(-54)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{79}}{2*16}=\frac{40-8\sqrt{79}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{79}}{2*16}=\frac{40+8\sqrt{79}}{32}
$