(M+5)(m-4)=(m-6)(m+1)

Solution for (M+5)(m-4)=(m-6)(m+1) equation:

(+5)(M-4)=(M-6)(M+1)

We move all terms to the left:

(+5)(M-4)-((M-6)(M+1))=0

We add all the numbers together, and all the variables

5(M-4)-((M-6)(M+1))=0

We multiply parentheses

5M-((M-6)(M+1))-20=0

We multiply parentheses ..

-((+M^2+M-6M-6))+5M-20=0


We calculate terms in parentheses: -((+M^2+M-6M-6)), so:

(+M^2+M-6M-6)

We get rid of parentheses

M^2+M-6M-6

We add all the numbers together, and all the variables

M^2-5M-6

Back to the equation:

-(M^2-5M-6)


We add all the numbers together, and all the variables

5M-(M^2-5M-6)-20=0

We get rid of parentheses

-M^2+5M+5M+6-20=0

We add all the numbers together, and all the variables

-1M^2+10M-14=0

a = -1; b = 10; c = -14;
Δ = b2-4ac
Δ = 102-4·(-1)·(-14)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{11}}{2*-1}=\frac{-10-2\sqrt{11}}{-2}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{11}}{2*-1}=\frac{-10+2\sqrt{11}}{-2}
$