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Type in the greater of the two values that satisfies the equation. Input your answer correct to three decimal places.

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\\(x\\) = [[0]]

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Type in the lesser of the two values that satisfies the equation. Input your answer correct to three decimal places.

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\\(x\\) = [[1]]

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The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

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\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

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In this example  \\(a=\\var{a1},\\,\\,\\,b=\\var{b1}\\)  and  \\(c=\\var{c1}\\)

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\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\var{b1}^2-4\\times\\var{a1}\\times\\var{c1}}}{2\\times\\var{a1}}\\)

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\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\simplify{{b1}^2-4*{a1}*{c1}}}}{\\simplify{2*{a1}}}\\)

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\\(x=\\simplify{(-{b1}+ ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)   or   \\(x=\\simplify{(-{b1}- ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)

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Solving quadratic equations using a formula,

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There are two values that satisfy the quadratic function below when  \\(y=\\var{c1}\\):

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\\(y=\\var{a1}x^2+\\var{b1}x\\)

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