(2/3)(9b-27)=3b

Solution for (2/3)(9b-27)=3b equation:

(2/3)(9b-27)=3b

We move all terms to the left:

(2/3)(9b-27)-(3b)=0


Domain of the equation: 3)(9b-27)!=0

b∈R


We add all the numbers together, and all the variables

(+2/3)(9b-27)-3b=0

We add all the numbers together, and all the variables

-3b+(+2/3)(9b-27)=0

We multiply parentheses ..

(+18b^2+2/3*-27)-3b=0

We multiply all the terms by the denominator


(+18b^2+2-3b*3*-27)=0

We get rid of parentheses

18b^2-3b*3*+2-27=0

We add all the numbers together, and all the variables

18b^2-3b*3*-25=0

Wy multiply elements

18b^2-9b^2-25=0

We add all the numbers together, and all the variables

9b^2-25=0

a = 9; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·9·(-25)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*9}=\frac{-30}{18}
=-1+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*9}=\frac{30}{18}
=1+2/3
$