(x-5)(x+3)(x+2)(x-6)=48

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Solution for (x-5)(x+3)(x+2)(x-6)=48 equation: 



(x-5)(x+3)(x+2)(x-6)=48

We move all terms to the left:

(x-5)(x+3)(x+2)(x-6)-(48)=0

We multiply parentheses ..

(+x^2+3x-5x-15)(x+2)(x-6)-48=0

We multiply parentheses ..

(+x^2+3x-5x-15)(+x^2-6x+2x-12)-48=0

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

(+x^2+3x-5x-15)(+x^2-6x+2x-12)=48

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