Answer:
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=3,\:x=-\frac{1}{2}[/tex]
Step-by-step explanation:
considering the equation
[tex]2x^2\:-\:4x\:-\:3\:=\:x[/tex]
solving
[tex]2x^2\:-\:4x\:-\:3\:=\:x[/tex]
[tex]2x^2-4x-3-x=x-x[/tex]
[tex]2x^2-5x-3=0[/tex]
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\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}[/tex]
solving
[tex]x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}[/tex]
[tex]x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}[/tex]
[tex]x=\frac{5+\sqrt{49}}{2\cdot \:2}[/tex]
[tex]x=\frac{5+7}{4}[/tex]
[tex]x=3[/tex]
also solving
[tex]x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}[/tex]
[tex]x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}[/tex]
[tex]x=\frac{5-\sqrt{49}}{4}[/tex]
[tex]x=-\frac{2}{4}[/tex]
[tex]x=-\frac{1}{2}[/tex]
Therefore,
[tex]\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}[/tex]
[tex]x=3,\:x=-\frac{1}{2}[/tex]
