12=-16t^2+96+12

Solution for 12=-16t^2+96+12 equation:

12=-16t^2+96+12

We move all terms to the left:

12-(-16t^2+96+12)=0

We get rid of parentheses

16t^2-96-12+12=0

We add all the numbers together, and all the variables

16t^2-96=0

a = 16; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·16·(-96)
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{6}}{2*16}=\frac{0-32\sqrt{6}}{32}
=-\frac{32\sqrt{6}}{32}
=-\sqrt{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{6}}{2*16}=\frac{0+32\sqrt{6}}{32}
=\frac{32\sqrt{6}}{32}
=\sqrt{6}
$