12/1+q=1/3q+2

Solution for 12/1+q=1/3q+2 equation:

12/1+q=1/3q+2

We move all terms to the left:

12/1+q-(1/3q+2)=0


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

q∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

q-1/3q-2=0

We multiply all the terms by the denominator


q*3q-2*3q-1=0

Wy multiply elements

3q^2-6q-1=0

a = 3; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·3·(-1)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{3}}{2*3}=\frac{6-4\sqrt{3}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{3}}{2*3}=\frac{6+4\sqrt{3}}{6}
$