0=3t^2-22t-16

Solution for 0=3t^2-22t-16 equation:

0=3t^2-22t-16

We move all terms to the left:

0-(3t^2-22t-16)=0

We add all the numbers together, and all the variables

-(3t^2-22t-16)=0

We get rid of parentheses

-3t^2+22t+16=0

a = -3; b = 22; c = +16;
Δ = b2-4ac
Δ = 222-4·(-3)·16
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*-3}=\frac{-48}{-6}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*-3}=\frac{4}{-6}
=-2/3
$