(1/3)m+(2/5)m=1100

Solution for (1/3)m+(2/5)m=1100 equation:

(1/3)m+(2/5)m=1100

We move all terms to the left:

(1/3)m+(2/5)m-(1100)=0


Domain of the equation: 3)m!=0

m!=0/1

m!=0

m∈R



Domain of the equation: 5)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+1/3)m+(+2/5)m-1100=0

We multiply parentheses

m^2+2m^2-1100=0

We add all the numbers together, and all the variables

3m^2-1100=0

a = 3; b = 0; c = -1100;
Δ = b2-4ac
Δ = 02-4·3·(-1100)
Δ = 13200
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:

$\sqrt{\Delta}=\sqrt{13200}=\sqrt{400*33}=\sqrt{400}*\sqrt{33}=20\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{33}}{2*3}=\frac{0-20\sqrt{33}}{6}
=-\frac{20\sqrt{33}}{6}
=-\frac{10\sqrt{33}}{3}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{33}}{2*3}=\frac{0+20\sqrt{33}}{6}
=\frac{20\sqrt{33}}{6}
=\frac{10\sqrt{33}}{3}
$