2/3q=1/2q-3

Solution for 2/3q=1/2q-3 equation:

2/3q=1/2q-3

We move all terms to the left:

2/3q-(1/2q-3)=0


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R



Domain of the equation: 2q-3)!=0

q∈R


We get rid of parentheses

2/3q-1/2q+3=0

We calculate fractions


4q/6q^2+(-3q)/6q^2+3=0

We multiply all the terms by the denominator


4q+(-3q)+3*6q^2=0

Wy multiply elements

18q^2+4q+(-3q)=0

We get rid of parentheses

18q^2+4q-3q=0

We add all the numbers together, and all the variables

18q^2+q=0

a = 18; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·18·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*18}=\frac{-2}{36}
=-1/18
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*18}=\frac{0}{36}
=0
$