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

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

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

We move all terms to the left:

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


Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

12b^2-6b^2-30b+3=0

We add all the numbers together, and all the variables

6b^2-30b+3=0

a = 6; b = -30; c = +3;
Δ = b2-4ac
Δ = -302-4·6·3
Δ = 828
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:

$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{23}}{2*6}=\frac{30-6\sqrt{23}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{23}}{2*6}=\frac{30+6\sqrt{23}}{12}
$