x/9x-13=3x+41

Solution for x/9x-13=3x+41 equation:

x/9x-13=3x+41

We move all terms to the left:

x/9x-13-(3x+41)=0


Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R


We get rid of parentheses

x/9x-3x-41-13=0

We multiply all the terms by the denominator


x-3x*9x-41*9x-13*9x=0

Wy multiply elements

-27x^2+x-369x-117x=0

We add all the numbers together, and all the variables

-27x^2-485x=0

a = -27; b = -485; c = 0;
Δ = b2-4ac
Δ = -4852-4·(-27)·0
Δ = 235225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{235225}=485$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-485)-485}{2*-27}=\frac{0}{-54}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-485)+485}{2*-27}=\frac{970}{-54}
=-17+26/27
$