0=16t^2+72t+12

Solution for 0=16t^2+72t+12 equation:

0=16t^2+72t+12

We move all terms to the left:

0-(16t^2+72t+12)=0

We add all the numbers together, and all the variables

-(16t^2+72t+12)=0

We get rid of parentheses

-16t^2-72t-12=0

a = -16; b = -72; c = -12;
Δ = b2-4ac
Δ = -722-4·(-16)·(-12)
Δ = 4416
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:

$\sqrt{\Delta}=\sqrt{4416}=\sqrt{64*69}=\sqrt{64}*\sqrt{69}=8\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{69}}{2*-16}=\frac{72-8\sqrt{69}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{69}}{2*-16}=\frac{72+8\sqrt{69}}{-32}
$