0=16t^2+72t+3

Solution for 0=16t^2+72t+3 equation:

0=16t^2+72t+3

We move all terms to the left:

0-(16t^2+72t+3)=0

We add all the numbers together, and all the variables

-(16t^2+72t+3)=0

We get rid of parentheses

-16t^2-72t-3=0

a = -16; b = -72; c = -3;
Δ = b2-4ac
Δ = -722-4·(-16)·(-3)
Δ = 4992
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:

$\sqrt{\Delta}=\sqrt{4992}=\sqrt{64*78}=\sqrt{64}*\sqrt{78}=8\sqrt{78}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{78}}{2*-16}=\frac{72-8\sqrt{78}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{78}}{2*-16}=\frac{72+8\sqrt{78}}{-32}
$