(7x^2+3x)/(5x+9)=11

Solution for (7x^2+3x)/(5x+9)=11 equation:

(7x^2+3x)/(5x+9)=11

We move all terms to the left:

(7x^2+3x)/(5x+9)-(11)=0


Domain of the equation: (5x+9)!=0

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

5x!=-9

x!=-9/5

x!=-1+4/5

x∈R


We multiply all the terms by the denominator


(7x^2+3x)-11*(5x+9)=0

We multiply parentheses

(7x^2+3x)-55x-99=0

We get rid of parentheses

7x^2+3x-55x-99=0

We add all the numbers together, and all the variables

7x^2-52x-99=0

a = 7; b = -52; c = -99;
Δ = b2-4ac
Δ = -522-4·7·(-99)
Δ = 5476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5476}=74$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-74}{2*7}=\frac{-22}{14}
=-1+4/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+74}{2*7}=\frac{126}{14}
=9
$