(2/3)q=2q+12

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

(2/3)q=2q+12

We move all terms to the left:

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


Domain of the equation: 3)q!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

2q^2-(2q+12)=0

We get rid of parentheses

2q^2-2q-12=0

a = 2; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*2}=\frac{-8}{4}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*2}=\frac{12}{4}
=3
$