2/3b+4=-b+8

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

2/3b+4=-b+8

We move all terms to the left:

2/3b+4-(-b+8)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R


We add all the numbers together, and all the variables

2/3b-(-1b+8)+4=0

We get rid of parentheses

2/3b+1b-8+4=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

3b^2-12b+2=0

a = 3; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·3·2
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{30}}{2*3}=\frac{12-2\sqrt{30}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{30}}{2*3}=\frac{12+2\sqrt{30}}{6}
$