0=-16t^2+40+24

Solution for 0=-16t^2+40+24 equation:

0=-16t^2+40+24

We move all terms to the left:

0-(-16t^2+40+24)=0

We add all the numbers together, and all the variables

-(-16t^2+40+24)=0

We get rid of parentheses

16t^2-40-24=0

We add all the numbers together, and all the variables

16t^2-64=0

a = 16; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·16·(-64)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*16}=\frac{64}{32}
=2
$