(x+3)(x-5)=(x+1)(x-2)

Solution for (x+3)(x-5)=(x+1)(x-2) equation:

(x+3)(x-5)=(x+1)(x-2)

We move all terms to the left:

(x+3)(x-5)-((x+1)(x-2))=0

We multiply parentheses ..

(+x^2-5x+3x-15)-((x+1)(x-2))=0


We calculate terms in parentheses: -((x+1)(x-2)), so:

(x+1)(x-2)

We multiply parentheses ..

(+x^2-2x+x-2)

We get rid of parentheses

x^2-2x+x-2

We add all the numbers together, and all the variables

x^2-1x-2

Back to the equation:

-(x^2-1x-2)


We get rid of parentheses

x^2-x^2-5x+3x+1x-15+2=0

We add all the numbers together, and all the variables

-1x-13=0

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

-x=13

x=13/-1

x=-13