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

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

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

We move all terms to the left:

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


Domain of the equation: 2m!=0

m!=0/2

m!=0

m∈R



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

m∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

320m^2+m=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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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