(2/3)b+(1/3)b+1+b=15

Solution for (2/3)b+(1/3)b+1+b=15 equation:

(2/3)b+(1/3)b+1+b=15

We move all terms to the left:

(2/3)b+(1/3)b+1+b-(15)=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+(+1/3)b+b+1-15=0

We add all the numbers together, and all the variables

b+(+2/3)b+(+1/3)b-14=0

We multiply parentheses

2b^2+b^2+b-14=0

We add all the numbers together, and all the variables

3b^2+b-14=0

a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*3}=\frac{-14}{6}
=-2+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6}
=2
$