2/3b+1/2b=3/4

Solution for 2/3b+1/2b=3/4 equation:

2/3b+1/2b=3/4

We move all terms to the left:

2/3b+1/2b-(3/4)=0


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We add all the numbers together, and all the variables

2/3b+1/2b-(+3/4)=0

We get rid of parentheses

2/3b+1/2b-3/4=0

We calculate fractions


(-36b^2)/96b^2+64b/96b^2+48b/96b^2=0

We multiply all the terms by the denominator


(-36b^2)+64b+48b=0

We add all the numbers together, and all the variables

(-36b^2)+112b=0

We get rid of parentheses

-36b^2+112b=0

a = -36; b = 112; c = 0;
Δ = b2-4ac
Δ = 1122-4·(-36)·0
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-112}{2*-36}=\frac{-224}{-72}
=3+1/9
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+112}{2*-36}=\frac{0}{-72}
=0
$