0=-16t^2+240t+12

Solution for 0=-16t^2+240t+12 equation:

0=-16t^2+240t+12

We move all terms to the left:

0-(-16t^2+240t+12)=0

We add all the numbers together, and all the variables

-(-16t^2+240t+12)=0

We get rid of parentheses

16t^2-240t-12=0

a = 16; b = -240; c = -12;
Δ = b2-4ac
Δ = -2402-4·16·(-12)
Δ = 58368
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:

$\sqrt{\Delta}=\sqrt{58368}=\sqrt{1024*57}=\sqrt{1024}*\sqrt{57}=32\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-32\sqrt{57}}{2*16}=\frac{240-32\sqrt{57}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+32\sqrt{57}}{2*16}=\frac{240+32\sqrt{57}}{32}
$