4(3b-1)=9/b

Solution for 4(3b-1)=9/b equation:

4(3b-1)=9/b

We move all terms to the left:

4(3b-1)-(9/b)=0


Domain of the equation: b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

4(3b-1)-(+9/b)=0

We multiply parentheses

12b-(+9/b)-4=0

We get rid of parentheses

12b-9/b-4=0

We multiply all the terms by the denominator


12b*b-4*b-9=0

We add all the numbers together, and all the variables

-4b+12b*b-9=0

Wy multiply elements

12b^2-4b-9=0

a = 12; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·12·(-9)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{7}}{2*12}=\frac{4-8\sqrt{7}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{7}}{2*12}=\frac{4+8\sqrt{7}}{24}
$