1/3b+8=1,2b-4

Solution for 1/3b+8=1,2b-4 equation:

1/3b+8=1.2b-4

We move all terms to the left:

1/3b+8-(1.2b-4)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We get rid of parentheses

1/3b-1.2b+4+8=0

We multiply all the terms by the denominator


-(1.2b)*3b+4*3b+8*3b+1=0

We add all the numbers together, and all the variables

-(+1.2b)*3b+4*3b+8*3b+1=0

We multiply parentheses

-3b^2+4*3b+8*3b+1=0

Wy multiply elements

-3b^2+12b+24b+1=0

We add all the numbers together, and all the variables

-3b^2+36b+1=0

a = -3; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·(-3)·1
Δ = 1308
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:

$\sqrt{\Delta}=\sqrt{1308}=\sqrt{4*327}=\sqrt{4}*\sqrt{327}=2\sqrt{327}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{327}}{2*-3}=\frac{-36-2\sqrt{327}}{-6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{327}}{2*-3}=\frac{-36+2\sqrt{327}}{-6}
$