0=-16t^2-4t+12

Solution for 0=-16t^2-4t+12 equation:

0=-16t^2-4t+12

We move all terms to the left:

0-(-16t^2-4t+12)=0

We add all the numbers together, and all the variables

-(-16t^2-4t+12)=0

We get rid of parentheses

16t^2+4t-12=0

a = 16; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·16·(-12)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*16}=\frac{24}{32}
=3/4
$