2/3b+5=10-b

Solution for 2/3b+5=10-b equation:

2/3b+5=10-b

We move all terms to the left:

2/3b+5-(10-b)=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+10)+5=0

We get rid of parentheses

2/3b+1b-10+5=0

We multiply all the terms by the denominator


1b*3b-10*3b+5*3b+2=0

Wy multiply elements

3b^2-30b+15b+2=0

We add all the numbers together, and all the variables

3b^2-15b+2=0

a = 3; b = -15; c = +2;
Δ = b2-4ac
Δ = -152-4·3·2
Δ = 201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{201}}{2*3}=\frac{15-\sqrt{201}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{201}}{2*3}=\frac{15+\sqrt{201}}{6}
$