12=-16t^2+28+2

Solution for 12=-16t^2+28+2 equation:

12=-16t^2+28+2

We move all terms to the left:

12-(-16t^2+28+2)=0

We get rid of parentheses

16t^2-28-2+12=0

We add all the numbers together, and all the variables

16t^2-18=0

a = 16; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·16·(-18)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*16}=\frac{0-24\sqrt{2}}{32}
=-\frac{24\sqrt{2}}{32}
=-\frac{3\sqrt{2}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*16}=\frac{0+24\sqrt{2}}{32}
=\frac{24\sqrt{2}}{32}
=\frac{3\sqrt{2}}{4}
$