0=-16t^2+44t+12

Solution for 0=-16t^2+44t+12 equation:

0=-16t^2+44t+12

We move all terms to the left:

0-(-16t^2+44t+12)=0

We add all the numbers together, and all the variables

-(-16t^2+44t+12)=0

We get rid of parentheses

16t^2-44t-12=0

a = 16; b = -44; c = -12;
Δ = b2-4ac
Δ = -442-4·16·(-12)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-52}{2*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+52}{2*16}=\frac{96}{32}
=3
$