(2/3)b+5=20+b

Solution for (2/3)b+5=20+b equation:

(2/3)b+5=20+b

We move all terms to the left:

(2/3)b+5-(20+b)=0


Domain of the equation: 3)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+2/3)b-(b+20)+5=0

We multiply parentheses

2b^2-(b+20)+5=0

We get rid of parentheses

2b^2-b-20+5=0

We add all the numbers together, and all the variables

2b^2-1b-15=0

a = 2; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*2}=\frac{-10}{4}
=-2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*2}=\frac{12}{4}
=3
$