7/2x+20=1/3x-50

Solution for 7/2x+20=1/3x-50 equation:

7/2x+20=1/3x-50

We move all terms to the left:

7/2x+20-(1/3x-50)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



Domain of the equation: 3x-50)!=0

x∈R


We get rid of parentheses

7/2x-1/3x+50+20=0

We calculate fractions


21x/6x^2+(-2x)/6x^2+50+20=0

We add all the numbers together, and all the variables

21x/6x^2+(-2x)/6x^2+70=0

We multiply all the terms by the denominator


21x+(-2x)+70*6x^2=0

Wy multiply elements

420x^2+21x+(-2x)=0

We get rid of parentheses

420x^2+21x-2x=0

We add all the numbers together, and all the variables

420x^2+19x=0

a = 420; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·420·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*420}=\frac{-38}{840}
=-19/420
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*420}=\frac{0}{840}
=0
$