(4/9)m=3

Solution for (4/9)m=3 equation:

(4/9)m=3

We move all terms to the left:

(4/9)m-(3)=0


Domain of the equation: 9)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+4/9)m-3=0

We multiply parentheses

4m^2-3=0

a = 4; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·4·(-3)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*4}=\frac{0-4\sqrt{3}}{8}
=-\frac{4\sqrt{3}}{8}
=-\frac{\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*4}=\frac{0+4\sqrt{3}}{8}
=\frac{4\sqrt{3}}{8}
=\frac{\sqrt{3}}{2}
$