(x-2)(x-4)(x+3)(x+5)=120S

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Solution for (x-2)(x-4)(x+3)(x+5)=120S equation: 



(x-2)(x-4)(x+3)(x+5)=120

We move all terms to the left:

(x-2)(x-4)(x+3)(x+5)-(120)=0

We multiply parentheses ..

(+x^2-4x-2x+8)(x+3)(x+5)-120=0

We multiply parentheses ..

(+x^2-4x-2x+8)(+x^2+5x+3x+15)-120=0

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

(+x^2-4x-2x+8)(+x^2+5x+3x+15)=120

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