12=q+2/3q=

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

12=q+2/3q=

We move all terms to the left:

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


Domain of the equation: 3q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-q-2/3q+12=0

We multiply all the terms by the denominator


-q*3q+12*3q-2=0

Wy multiply elements

-3q^2+36q-2=0

a = -3; b = 36; c = -2;
Δ = b2-4ac
Δ = 362-4·(-3)·(-2)
Δ = 1272
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:

$\sqrt{\Delta}=\sqrt{1272}=\sqrt{4*318}=\sqrt{4}*\sqrt{318}=2\sqrt{318}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{318}}{2*-3}=\frac{-36-2\sqrt{318}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{318}}{2*-3}=\frac{-36+2\sqrt{318}}{-6}
$