q/3q+5+2q-5+2q-5=65

Solution for q/3q+5+2q-5+2q-5=65 equation:

q/3q+5+2q-5+2q-5=65

We move all terms to the left:

q/3q+5+2q-5+2q-5-(65)=0


Domain of the equation: 3q!=0

q!=0/3

q!=0

q∈R


We add all the numbers together, and all the variables

4q+q/3q-70=0

We multiply all the terms by the denominator


4q*3q+q-70*3q=0

We add all the numbers together, and all the variables

q+4q*3q-70*3q=0

Wy multiply elements

12q^2+q-210q=0

We add all the numbers together, and all the variables

12q^2-209q=0

a = 12; b = -209; c = 0;
Δ = b2-4ac
Δ = -2092-4·12·0
Δ = 43681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{43681}=209$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-209)-209}{2*12}=\frac{0}{24}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-209)+209}{2*12}=\frac{418}{24}
=17+5/12
$