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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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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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Solving quadratic equations using a formula,

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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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