40/20x=9/x+13

Solution for 40/20x=9/x+13 equation:

40/20x=9/x+13

We move all terms to the left:

40/20x-(9/x+13)=0


Domain of the equation: 20x!=0

x!=0/20

x!=0

x∈R



Domain of the equation: x+13)!=0

x∈R


We get rid of parentheses

40/20x-9/x-13=0

We calculate fractions


40x/20x^2+(-180x)/20x^2-13=0

We multiply all the terms by the denominator


40x+(-180x)-13*20x^2=0

Wy multiply elements

-260x^2+40x+(-180x)=0

We get rid of parentheses

-260x^2+40x-180x=0

We add all the numbers together, and all the variables

-260x^2-140x=0

a = -260; b = -140; c = 0;
Δ = b2-4ac
Δ = -1402-4·(-260)·0
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-140}{2*-260}=\frac{0}{-520}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+140}{2*-260}=\frac{280}{-520}
=-7/13
$