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The formula for solving a quadratic equation of the form \\(ax^2+bx+c=0\\) is given by
\n\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)
\nIn this example \\(a=\\var{a1},\\,\\,\\,b=\\var{b1}\\) and \\(c=\\var{c1}\\)
\n\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\var{b1}^2-4\\times\\var{a1}\\times\\var{c1}}}{2\\times\\var{a1}}\\)
\n\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\simplify{{b1}^2-4*{a1}*{c1}}}}{\\simplify{2*{a1}}}\\)
\n\\(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})}\\)
", "parts": [{"prompt": "Type in the greater of the two values that satisfies the equation. Input your answer correct to three decimal places.
\n\\(x\\) = [[0]]
\nType in the lesser of the two values that satisfies the equation. Input your answer correct to three decimal places.
\n\\(x\\) = [[1]]
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\n\\(\\var{a1}x^2+\\var{b1}x+\\var{c1}=0\\)
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", "licence": "Creative Commons Attribution 4.0 International"}, "contributors": [{"name": "Andrew Dunbar", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/770/"}]}]}], "contributors": [{"name": "Andrew Dunbar", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/770/"}]}