Question

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

Answer 1

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