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

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

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

We move all terms to the left:

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


Domain of the equation: 3b!=0

b!=0/3

b!=0

b∈R



Domain of the equation: 4b+2)!=0

b∈R


We get rid of parentheses

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

We calculate fractions


8b/12b^2+9b/12b^2-2+1=0

We add all the numbers together, and all the variables

8b/12b^2+9b/12b^2-1=0

We multiply all the terms by the denominator


8b+9b-1*12b^2=0

We add all the numbers together, and all the variables

17b-1*12b^2=0

Wy multiply elements

-12b^2+17b=0

a = -12; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-12)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-12}=\frac{-34}{-24}
=1+5/12
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-12}=\frac{0}{-24}
=0
$