40/5x-2=48/2x+7

Solution for 40/5x-2=48/2x+7 equation:

40/5x-2=48/2x+7

We move all terms to the left:

40/5x-2-(48/2x+7)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 2x+7)!=0

x∈R


We get rid of parentheses

40/5x-48/2x-7-2=0

We calculate fractions


80x/10x^2+(-240x)/10x^2-7-2=0

We add all the numbers together, and all the variables

80x/10x^2+(-240x)/10x^2-9=0

We multiply all the terms by the denominator


80x+(-240x)-9*10x^2=0

Wy multiply elements

-90x^2+80x+(-240x)=0

We get rid of parentheses

-90x^2+80x-240x=0

We add all the numbers together, and all the variables

-90x^2-160x=0

a = -90; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·(-90)·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-160}{2*-90}=\frac{0}{-180}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*-90}=\frac{320}{-180}
=-1+7/9
$