(M+3)(m)(m+3)=1

Solution for (M+3)(m)(m+3)=1 equation:

(+3)(M)(M+3)=1

We move all terms to the left:

(+3)(M)(M+3)-(1)=0

We add all the numbers together, and all the variables

3M(M+3)-1=0

We multiply parentheses

3M^2+9M-1=0

a = 3; b = 9; c = -1;
Δ = b2-4ac
Δ = 92-4·3·(-1)
Δ = 93
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{93}}{2*3}=\frac{-9-\sqrt{93}}{6}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{93}}{2*3}=\frac{-9+\sqrt{93}}{6}
$