(M+3)(m-5)=(m-2)(m+5)

Solution for (M+3)(m-5)=(m-2)(m+5) equation:

(+3)(M-5)=(M-2)(M+5)

We move all terms to the left:

(+3)(M-5)-((M-2)(M+5))=0

We add all the numbers together, and all the variables

3(M-5)-((M-2)(M+5))=0

We multiply parentheses

3M-((M-2)(M+5))-15=0

We multiply parentheses ..

-((+M^2+5M-2M-10))+3M-15=0


We calculate terms in parentheses: -((+M^2+5M-2M-10)), so:

(+M^2+5M-2M-10)

We get rid of parentheses

M^2+5M-2M-10

We add all the numbers together, and all the variables

M^2+3M-10

Back to the equation:

-(M^2+3M-10)


We add all the numbers together, and all the variables

3M-(M^2+3M-10)-15=0

We get rid of parentheses

-M^2+3M-3M+10-15=0

We add all the numbers together, and all the variables

-1M^2-5=0

a = -1; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·(-1)·(-5)
Δ = -20
Delta is less than zero, so there is no solution for the equation