0=-4b+39-3b^2

Solution for 0=-4b+39-3b^2 equation:

0=-4b+39-3b^2

We move all terms to the left:

0-(-4b+39-3b^2)=0

We add all the numbers together, and all the variables

-(-4b+39-3b^2)=0

We get rid of parentheses

3b^2+4b-39=0

a = 3; b = 4; c = -39;
Δ = b2-4ac
Δ = 42-4·3·(-39)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-22}{2*3}=\frac{-26}{6}
=-4+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+22}{2*3}=\frac{18}{6}
=3
$