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

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

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

We move all terms to the left:

/3Q+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

2Q+/3Q-65=0

We multiply all the terms by the denominator


2Q*3Q-65*3Q+=0

We add all the numbers together, and all the variables

2Q*3Q-65*3Q=0

Wy multiply elements

6Q^2-195Q=0

a = 6; b = -195; c = 0;
Δ = b2-4ac
Δ = -1952-4·6·0
Δ = 38025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{38025}=195$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-195)-195}{2*6}=\frac{0}{12}
=0
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-195)+195}{2*6}=\frac{390}{12}
=32+1/2
$