3/2x+12=7/5x-20

Solution for 3/2x+12=7/5x-20 equation:

3/2x+12=7/5x-20

We move all terms to the left:

3/2x+12-(7/5x-20)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



Domain of the equation: 5x-20)!=0

x∈R


We get rid of parentheses

3/2x-7/5x+20+12=0

We calculate fractions


15x/10x^2+(-14x)/10x^2+20+12=0

We add all the numbers together, and all the variables

15x/10x^2+(-14x)/10x^2+32=0

We multiply all the terms by the denominator


15x+(-14x)+32*10x^2=0

Wy multiply elements

320x^2+15x+(-14x)=0

We get rid of parentheses

320x^2+15x-14x=0

We add all the numbers together, and all the variables

320x^2+x=0

a = 320; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·320·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*320}=\frac{-2}{640}
=-1/320
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*320}=\frac{0}{640}
=0
$