Why can't you use the zero-product property to solve (x-1)(x+3)=1?

If ( x - 1 ) * ( x + 3 ) = 0 => x = 1 or x = -3;
But, (x-1)(x+3)=1 => x^2 + 3x -x - 3 = 1 => x^2 + 2x - 4 = 0;the discriminant is 2^2 - 4 * 1 * ( -4 ) = 20 => the solutions are x1 = 
( - 2 + ) / 2 ; x2 = - 2 -  ) / 2;

Igotchu on that if you still need it

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