(3/4)m+6=18

Solution for (3/4)m+6=18 equation:

(3/4)m+6=18

We move all terms to the left:

(3/4)m+6-(18)=0


Domain of the equation: 4)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+3/4)m+6-18=0

We add all the numbers together, and all the variables

(+3/4)m-12=0

We multiply parentheses

3m^2-12=0

a = 3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·3·(-12)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*3}=\frac{-12}{6}
=-2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*3}=\frac{12}{6}
=2
$