3m2/5;m=2/3

Solution for 3m2/5;m=2/3 equation:

3m^2/5m=2/3

We move all terms to the left:

3m^2/5m-(2/3)=0


Domain of the equation: 5m!=0

m!=0/5

m!=0

m∈R


We add all the numbers together, and all the variables

3m^2/5m-(+2/3)=0

We get rid of parentheses

3m^2/5m-2/3=0

We calculate fractions


9m^2/15m+(-10m)/15m=0

We multiply all the terms by the denominator


9m^2+(-10m)=0

We get rid of parentheses

9m^2-10m=0

a = 9; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·9·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*9}=\frac{0}{18}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*9}=\frac{20}{18}
=1+1/9
$