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What are the zeros of the quadratic function f(x) = 6x- + 12x-7?
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[tex]~~~~~6x^2 +12x -7 =0\\\\\implies x^2 +2x - \dfrac 76=0\\\\\implies x^2+2x = \dfrac76\\\\\implies x^2 +2x +1 = \dfrac 76 +1\\\\\implies (x+1)^2 = \dfrac{13}6\\\\\implies x+1 = \pm \sqrt{ \dfrac{13}6}\\\\\implies x = -1 \pm\sqrt{\dfrac{13} 6}\\\\\text{Hence,}~ x = -1 +\sqrt{\dfrac{13} 6}~~ \text{and}~~ x = -1-\sqrt{\dfrac{13} 6}[/tex]
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Number 1

Step-by-step explanation:

Vertex is (−1,−13)
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