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  1. L

    Therefore, the roots of the given equation are $x=-4$ and $x=2$.

    Solve (x+3)^2 = 4x+17 where did i go wrong? (x+3)(x+3 )= 4x+17 x^2 + 3x + 3x + 9 = 4x+17 x^2 + 6x + 9 = 4x + 17 x^2 + 6x + 9 - 4x - 17 = 0 x^2 + 2x - 8 = 0 (x-2)(x+4) <-- USING THE CROSS METHOD x= -2, 4 is my cross method working incorrect or something? do explain my mistake
    • led
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    • Quadratic Quadratic equation
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    • Forum: General Math
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