(1/3)m+(4/3)=6

Solution for (1/3)m+(4/3)=6 equation:

(1/3)m+(4/3)=6

We move all terms to the left:

(1/3)m+(4/3)-(6)=0


Domain of the equation: 3)m!=0

m!=0/1

m!=0

m∈R


determiningTheFunctionDomain
(1/3)m-6+(4/3)=0

We add all the numbers together, and all the variables

(+1/3)m-6+(+4/3)=0

We multiply parentheses

m^2-6+(+4/3)=0

We get rid of parentheses

m^2-6+4/3=0

We multiply all the terms by the denominator


m^2*3+4-6*3=0

We add all the numbers together, and all the variables

m^2*3-14=0

Wy multiply elements

3m^2-14=0

a = 3; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·3·(-14)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*3}=\frac{0-2\sqrt{42}}{6}
=-\frac{2\sqrt{42}}{6}
=-\frac{\sqrt{42}}{3}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*3}=\frac{0+2\sqrt{42}}{6}
=\frac{2\sqrt{42}}{6}
=\frac{\sqrt{42}}{3}
$