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The following equation can be converted into a quadratic equation:

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\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

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Solving quadratic equations using a formula

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\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

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We clear the fraction in the equation by multiplying across by \\(x\\)

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\\(\\var{a1}x^2+\\simplify{{a1}*{b1}*{c1}}=\\simplify{{a1}*({b1}+{c1})}x\\)

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Bringing all the terms to the left hand side and putting them in order of their powers of \\(x\\) gives

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\\(\\var{a1}x^2-\\simplify{{a1}*({b1}+{c1})}x+\\simplify{{a1}*{b1}*{c1}}=0\\)

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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=\\simplify{+-{a1}*({b1}+{c1})}\\)  and  \\(c=\\simplify{{a1}*{b1}*{c1}}\\)

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\\(x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\\)

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

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

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Type in the greater of the two values that satisfies the equation. 

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\\(x = \\) [[0]]

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Type in the lesser of the two values that satisfies the equation. 

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\\(x = \\) [[1]]

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