0=-12t^2+16

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

0=-12t^2+16

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

12t^2-16=0

a = 12; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·12·(-16)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*12}=\frac{0-16\sqrt{3}}{24}
=-\frac{16\sqrt{3}}{24}
=-\frac{2\sqrt{3}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*12}=\frac{0+16\sqrt{3}}{24}
=\frac{16\sqrt{3}}{24}
=\frac{2\sqrt{3}}{3}
$