(2/3)m=2/5

Solution for (2/3)m=2/5 equation:

(2/3)m=2/5

We move all terms to the left:

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


Domain of the equation: 3)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

2m^2-2/5=0

We multiply all the terms by the denominator


2m^2*5-2=0

Wy multiply elements

10m^2-2=0

a = 10; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·10·(-2)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*10}=\frac{0-4\sqrt{5}}{20}
=-\frac{4\sqrt{5}}{20}
=-\frac{\sqrt{5}}{5}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*10}=\frac{0+4\sqrt{5}}{20}
=\frac{4\sqrt{5}}{20}
=\frac{\sqrt{5}}{5}
$