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$ax^2+bx+c=0$,

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the quadratic formula (which itself is a result of completing the square) is the solution

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$x=\\displaystyle{\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}$.

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For our quadratic $\\simplify{{scoeff}x^2+{lcoeff}x+{ccoeff}=0}$ we have $a=\\var{scoeff}$, $b=\\var{lcoeff}$ and $c=\\var{ccoeff}$, which gives us:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $x$ $=$ $\\displaystyle{\\frac{-(\\var{lcoeff})\\pm\\sqrt{(\\var{lcoeff})^2-4(\\var{scoeff})(\\var{ccoeff})}}{2(\\var{scoeff})}}$ $=$ $\\displaystyle{\\frac{\\var{-lcoeff}\\pm\\sqrt{\\var{lcoeff^2}-(\\var{4*scoeff*ccoeff})}}{\\var{2*scoeff}}}$ $=$ $\\displaystyle{\\frac{\\var{-lcoeff}\\pm\\sqrt{\\var{disc}}}{\\var{2*scoeff}}}$ $=$ $\\displaystyle{\\frac{\\var{-lcoeff}\\pm\\var{lengthdet}}{\\var{2*scoeff}}}$ $=$ $\\displaystyle{\\frac{\\var{-lcoeff-lengthdet}}{\\var{2*scoeff}},\\,\\,\\frac{\\var{-lcoeff+lengthdet}}{\\var{2*scoeff}}}$ $=$ $\\displaystyle{\\simplify{({-lcoeff}-{sqrt(disc)})/(2*{scoeff})},\\,\\,\\simplify{({-lcoeff}+{sqrt(disc)})/(2*{scoeff})}}$
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$\\simplify{{scoeff}x^2+{lcoeff}x+{ccoeff}=0}$.

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$x=$ [[0]], [[1]]

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Note: Put the smallest value (the one with the negative in front of the square root) in the first gap.

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