0=-16t^2+72t+4

Solution for 0=-16t^2+72t+4 equation:

0=-16t^2+72t+4

We move all terms to the left:

0-(-16t^2+72t+4)=0

We add all the numbers together, and all the variables

-(-16t^2+72t+4)=0

We get rid of parentheses

16t^2-72t-4=0

a = 16; b = -72; c = -4;
Δ = b2-4ac
Δ = -722-4·16·(-4)
Δ = 5440
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:

$\sqrt{\Delta}=\sqrt{5440}=\sqrt{64*85}=\sqrt{64}*\sqrt{85}=8\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{85}}{2*16}=\frac{72-8\sqrt{85}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{85}}{2*16}=\frac{72+8\sqrt{85}}{32}
$