Answer:
[tex]x=\frac{-1+\sqrt{6}}{5},\:x=-\frac{1+\sqrt{6}}{5}[/tex]
Step-by-step explanation:
Method 1 ; Completing the square method
[tex]5x^2+2x=1\\\mathrm{Divide\:both\:sides\:by\:}5\\\frac{5x^2+2x}{5}=\frac{1}{5}\\\\x^2+\frac{2x}{5}=\frac{1}{5}\\\\\mathrm{Solve\:by\:completing\:the\:square}\\\\Add\:the\:square\:of\:the\:half\:f\:the\:coefficient\\\:of\:x\:to\:both\:sides\\\\\frac{2}{5} \times \frac{1}{2} = (\frac{1}{5} )^2\\\\x^2+\frac{2}{5} x +(\frac{1}{5} )^2 =\frac{1}{5} +(\frac{1}{5} )^2\\\\x^2+ \frac{2}{5} x +(\frac{1}{5} )^2 =\frac{1}{5} +\frac{1}{25}\\\\x^2+ \frac{2}{5} x +(\frac{1}{5} )^2 = \frac{6}{25} \\[/tex]
[tex](x + \frac{1}{5} )^2 = \frac{6}{25} \\Square\:root\:both\:sides\:of\:the\:equation\\\sqrt{(x+\frac{1}{5})^2 } = \sqrt{\frac{6}{25} } \\\\x + \frac{1}{5} =\frac{\sqrt{6} }{5} \\\\x = -\frac{1}{5}\±\frac{\sqrt{6} }{5} \\\\x = \frac{-1+\sqrt{6} }{5} \\\\x = \frac{-1-\sqrt{6} }{5} \\\\\\x = \frac{\sqrt{6} -1}{2}\\\\x = \frac{-\sqrt{6} -1}{2}[/tex]
Method 2 ; Quadratic formula
[tex]\frac{x}{y} 5x^2+2x=1\\5x^2+2x-1=0\\\\a=5,\:b=2,\:c=-1\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x =\frac{-2\pm \sqrt{2^2-4\times\:5\left(-1\right)}}{2\times\:5}\\\\x = \frac{-2\pm \sqrt{4+20} }{10} \\\\x = \frac{-2\pm \sqrt{25} }{10} \\\\\mathrm{Cancel\:the\:common\:factor:}\:2\\x = \frac{-1\pm\sqrt{6} }{5} \\[/tex]
