6/(m+3)+21=7m

Solution for 6/(m+3)+21=7m equation:

6/(m+3)+21=7m

We move all terms to the left:

6/(m+3)+21-(7m)=0


Domain of the equation: (m+3)!=0

We move all terms containing m to the left, all other terms to the right

m!=-3

m∈R


We add all the numbers together, and all the variables

-7m+6/(m+3)+21=0

We multiply all the terms by the denominator


-7m*(m+3)+21*(m+3)+6=0

We multiply parentheses

-7m^2-21m+21m+63+6=0

We add all the numbers together, and all the variables

-7m^2+69=0

a = -7; b = 0; c = +69;
Δ = b2-4ac
Δ = 02-4·(-7)·69
Δ = 1932
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:

$\sqrt{\Delta}=\sqrt{1932}=\sqrt{4*483}=\sqrt{4}*\sqrt{483}=2\sqrt{483}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{483}}{2*-7}=\frac{0-2\sqrt{483}}{-14}
=-\frac{2\sqrt{483}}{-14}
=-\frac{\sqrt{483}}{-7}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{483}}{2*-7}=\frac{0+2\sqrt{483}}{-14}
=\frac{2\sqrt{483}}{-14}
=\frac{\sqrt{483}}{-7}
$