0=-16t^2+72t+2.4

Solution for 0=-16t^2+72t+2.4 equation:

0=-16t^2+72t+2.4

We move all terms to the left:

0-(-16t^2+72t+2.4)=0

We add all the numbers together, and all the variables

-(-16t^2+72t+2.4)=0

We get rid of parentheses

16t^2-72t-2.4=0

a = 16; b = -72; c = -2.4;
Δ = b2-4ac
Δ = -722-4·16·(-2.4)
Δ = 5337.6
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{-(-72)-\sqrt{5337.6}}{2*16}=\frac{72-\sqrt{5337.6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+\sqrt{5337.6}}{2*16}=\frac{72+\sqrt{5337.6}}{32}
$