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Pavel [41]
3 years ago
14

Which of the following are solutions to the equation below?

Mathematics
1 answer:
sladkih [1.3K]3 years ago
7 0

Answer:

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

                  x=3,\:x=-\frac{1}{2}

Step-by-step explanation:

considering the equation

2x^2\:-\:4x\:-\:3\:=\:x

solving

2x^2\:-\:4x\:-\:3\:=\:x

2x^2-4x-3-x=x-x

2x^2-5x-3=0

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

solving

x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5+\sqrt{49}}{2\cdot \:2}

x=\frac{5+7}{4}

x=3

also solving

x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5-\sqrt{49}}{4}

x=-\frac{2}{4}

x=-\frac{1}{2}

Therefore,

                 \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

                  x=3,\:x=-\frac{1}{2}

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