(x-2)(x+3)=(x+4)(x-5)

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Solution for (x-2)(x+3)=(x+4)(x-5) equation: 



(x-2)(x+3)=(x+4)(x-5)

We move all terms to the left:

(x-2)(x+3)-((x+4)(x-5))=0

We multiply parentheses ..

(+x^2+3x-2x-6)-((x+4)(x-5))=0



We calculate terms in parentheses: -((x+4)(x-5)), so:

(x+4)(x-5)

We multiply parentheses ..

(+x^2-5x+4x-20)

We get rid of parentheses

x^2-5x+4x-20

We add all the numbers together, and all the variables

x^2-1x-20

Back to the equation:

-(x^2-1x-20)



We get rid of parentheses

x^2-x^2+3x-2x+1x-6+20=0

We add all the numbers together, and all the variables

2x+14=0

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

2x=-14

x=-14/2

x=-7

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