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Mathematics 63 Online
OpenStudy (anonymous):

2x^2-7x+3 solve by completing the square.

OpenStudy (campbell_st):

divide every term by 2 \[x^2 + \frac{7}{2} x +\frac{3}{2}\] halve the coefficient of x \[\frac{7}{2} \div 2 = \frac{7}{4}\] then its \[x^2 + \frac{7}{2}x +( \frac{7}{4})^2 + \frac{3}{2} - (\frac{7}{4})^2=0\] which is \[(x + \frac{7}{4})^2 = \frac{25}{4}\] take the square root of both sides \[x + \frac{7}{4} = \pm \frac{5}{2}\] I'll leave you to solve for x

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