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Use the quadratic formula to solve the following quadratic:
\n$\\simplify{{scoeff}x^2+{lcoeff}x+{ccoeff}=0}$.
\n\n$x=$ [[0]], [[1]]
\n", "stepsPenalty": "2", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given the quadratic
\n$ax^2+bx+c=0$,
\nthe quadratic formula (which itself is a result of completing the square) is the solution
\n$x=\\displaystyle{\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}$.
\n\nFor 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$\\begin{align*}x &=\\frac{-(\\var{lcoeff})\\pm\\sqrt{(\\var{lcoeff})^2-4(\\var{scoeff})(\\var{ccoeff})}}{2(\\var{scoeff})}\\\\
&=\\frac{\\var{-lcoeff}\\pm\\sqrt{\\var{lcoeff^2}-(\\var{4*scoeff*ccoeff})}}{\\var{2*scoeff}}\\\\
&= \\frac{\\var{-lcoeff}\\pm\\sqrt{\\var{disc}}}{\\var{2*scoeff}}\\\\
&= \\frac{\\var{-lcoeff}\\pm\\var{lengthdet}}{\\var{2*scoeff}}\\\\
&= \\frac{\\var{-lcoeff-lengthdet}}{\\var{2*scoeff}},\\,\\,\\frac{\\var{-lcoeff+lengthdet}}{\\var{2*scoeff}}\\\\
&=\\simplify{({-lcoeff}-{sqrt(disc)})/(2*{scoeff})},\\,\\,\\simplify{({-lcoeff}+{sqrt(disc)})/(2*{scoeff})} \\end{align*}$