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

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

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

We move all terms to the left:

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


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



Domain of the equation: 5x-2)!=0

x∈R


We get rid of parentheses

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

We calculate fractions


200x/10x^2+(-96x)/10x^2+2+7=0

We add all the numbers together, and all the variables

200x/10x^2+(-96x)/10x^2+9=0

We multiply all the terms by the denominator


200x+(-96x)+9*10x^2=0

Wy multiply elements

90x^2+200x+(-96x)=0

We get rid of parentheses

90x^2+200x-96x=0

We add all the numbers together, and all the variables

90x^2+104x=0

a = 90; b = 104; c = 0;
Δ = b2-4ac
Δ = 1042-4·90·0
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-104}{2*90}=\frac{-208}{180}
=-1+7/45
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+104}{2*90}=\frac{0}{180}
=0
$