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

Question

campbell_st · Answer

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