40/20x=9x+13

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

40/20x=9x+13

We move all terms to the left:

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


Domain of the equation: 20x!=0

x!=0/20

x!=0

x∈R


We get rid of parentheses

40/20x-9x-13=0

We multiply all the terms by the denominator


-9x*20x-13*20x+40=0

Wy multiply elements

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

a = -180; b = -260; c = +40;
Δ = b2-4ac
Δ = -2602-4·(-180)·40
Δ = 96400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96400}=\sqrt{400*241}=\sqrt{400}*\sqrt{241}=20\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-260)-20\sqrt{241}}{2*-180}=\frac{260-20\sqrt{241}}{-360}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-260)+20\sqrt{241}}{2*-180}=\frac{260+20\sqrt{241}}{-360}
$