(4/15)m+(1/5)=1

Solution for (4/15)m+(1/5)=1 equation:

(4/15)m+(1/5)=1

We move all terms to the left:

(4/15)m+(1/5)-(1)=0


Domain of the equation: 15)m!=0

m!=0/1

m!=0

m∈R


determiningTheFunctionDomain
(4/15)m-1+(1/5)=0

We add all the numbers together, and all the variables

(+4/15)m-1+(+1/5)=0

We multiply parentheses

4m^2-1+(+1/5)=0

We get rid of parentheses

4m^2-1+1/5=0

We multiply all the terms by the denominator


4m^2*5+1-1*5=0

We add all the numbers together, and all the variables

4m^2*5-4=0

Wy multiply elements

20m^2-4=0

a = 20; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·20·(-4)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*20}=\frac{0-8\sqrt{5}}{40}
=-\frac{8\sqrt{5}}{40}
=-\frac{\sqrt{5}}{5}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*20}=\frac{0+8\sqrt{5}}{40}
=\frac{8\sqrt{5}}{40}
=\frac{\sqrt{5}}{5}
$