(M+2)(m-2)=(2m+1)(2m-1)

Solution for (M+2)(m-2)=(2m+1)(2m-1) equation:

(+2)(M-2)=(2M+1)(2M-1)

We move all terms to the left:

(+2)(M-2)-((2M+1)(2M-1))=0

We add all the numbers together, and all the variables

2(M-2)-((2M+1)(2M-1))=0

We use the square of the difference formula

4M^2+2(M-2)+1=0

We multiply parentheses

4M^2+2M-4+1=0

We add all the numbers together, and all the variables

4M^2+2M-3=0

a = 4; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·4·(-3)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*4}=\frac{-2-2\sqrt{13}}{8}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*4}=\frac{-2+2\sqrt{13}}{8}
$