(3/2)b+6+(1/2)b=15+2b

Solution for (3/2)b+6+(1/2)b=15+2b equation:

(3/2)b+6+(1/2)b=15+2b

We move all terms to the left:

(3/2)b+6+(1/2)b-(15+2b)=0


Domain of the equation: 2)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+3/2)b+(+1/2)b-(2b+15)+6=0

We multiply parentheses

3b^2+b^2-(2b+15)+6=0

We get rid of parentheses

3b^2+b^2-2b-15+6=0

We add all the numbers together, and all the variables

4b^2-2b-9=0

a = 4; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·4·(-9)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{37}}{2*4}=\frac{2-2\sqrt{37}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{37}}{2*4}=\frac{2+2\sqrt{37}}{8}
$