(3x-9)/(x^2-9)=7

Solution for (3x-9)/(x^2-9)=7 equation:

(3x-9)/(x^2-9)=7

We move all terms to the left:

(3x-9)/(x^2-9)-(7)=0


Domain of the equation: (x^2-9)!=0

We move all terms containing x to the left, all other terms to the right

x^2!=9

x^2!=9/

x^2!=√1/0

x!=1

x∈R


We multiply all the terms by the denominator


(3x-9)-7*(x^2-9)=0

We multiply parentheses

-7x^2+(3x-9)+63=0

We get rid of parentheses

-7x^2+3x-9+63=0

We add all the numbers together, and all the variables

-7x^2+3x+54=0

a = -7; b = 3; c = +54;
Δ = b2-4ac
Δ = 32-4·(-7)·54
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-39}{2*-7}=\frac{-42}{-14}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+39}{2*-7}=\frac{36}{-14}
=-2+4/7
$