7(2m-1)-3m/5m=6/5(4-3)

Solution for 7(2m-1)-3m/5m=6/5(4-3) equation:

7(2m-1)-3m/5m=6/5(4-3)

We move all terms to the left:

7(2m-1)-3m/5m-(6/5(4-3))=0


Domain of the equation: 5m!=0

m!=0/5

m!=0

m∈R


We add all the numbers together, and all the variables

7(2m-1)-3m/5m-(6/51)=0

We multiply parentheses

14m-3m/5m-7-(6/51)=0

We get rid of parentheses

14m-3m/5m-7-6/51=0

We calculate fractions


14m+(-153m)/255m+(-30m)/255m-7=0

We multiply all the terms by the denominator


14m*255m+(-153m)+(-30m)-7*255m=0

Wy multiply elements

3570m^2+(-153m)+(-30m)-1785m=0

We get rid of parentheses

3570m^2-153m-30m-1785m=0

We add all the numbers together, and all the variables

3570m^2-1968m=0

a = 3570; b = -1968; c = 0;
Δ = b2-4ac
Δ = -19682-4·3570·0
Δ = 3873024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3873024}=1968$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1968)-1968}{2*3570}=\frac{0}{7140}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1968)+1968}{2*3570}=\frac{3936}{7140}
=328/595
$