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

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

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

We move all terms to the left:

0-(-3b^2+22b+16)=0

We add all the numbers together, and all the variables

-(-3b^2+22b+16)=0

We get rid of parentheses

3b^2-22b-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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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