40x/20x=9x+x+13

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

40x/20x=9x+x+13

We move all terms to the left:

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


Domain of the equation: 20x!=0

x!=0/20

x!=0

x∈R


We add all the numbers together, and all the variables

40x/20x-(10x+13)=0

We get rid of parentheses

40x/20x-10x-13=0

We multiply all the terms by the denominator


40x-10x*20x-13*20x=0

Wy multiply elements

-200x^2+40x-260x=0

We add all the numbers together, and all the variables

-200x^2-220x=0

a = -200; b = -220; c = 0;
Δ = b2-4ac
Δ = -2202-4·(-200)·0
Δ = 48400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{48400}=220$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-220)-220}{2*-200}=\frac{0}{-400}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-220)+220}{2*-200}=\frac{440}{-400}
=-1+1/10
$