(7x^2+3x)/5x+9=12

Solution for (7x^2+3x)/5x+9=12 equation:

(7x^2+3x)/5x+9=12

We move all terms to the left:

(7x^2+3x)/5x+9-(12)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We add all the numbers together, and all the variables

(7x^2+3x)/5x-3=0

We multiply all the terms by the denominator


(7x^2+3x)-3*5x=0

Wy multiply elements

(7x^2+3x)-15x=0

We get rid of parentheses

7x^2+3x-15x=0

We add all the numbers together, and all the variables

7x^2-12x=0

a = 7; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·7·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*7}=\frac{0}{14}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*7}=\frac{24}{14}
=1+5/7
$