3/2b+9=3b

Solution for 3/2b+9=3b equation:

3/2b+9=3b

We move all terms to the left:

3/2b+9-(3b)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We add all the numbers together, and all the variables

-3b+3/2b+9=0

We multiply all the terms by the denominator


-3b*2b+9*2b+3=0

Wy multiply elements

-6b^2+18b+3=0

a = -6; b = 18; c = +3;
Δ = b2-4ac
Δ = 182-4·(-6)·3
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{11}}{2*-6}=\frac{-18-6\sqrt{11}}{-12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{11}}{2*-6}=\frac{-18+6\sqrt{11}}{-12}
$