Solve $u^2-7u-4=0$ by completing the square, if possible.

A) $y=\frac{\sqrt{65}}{2},y=-\frac{\sqrt{65}}{2}$
B) $y=\frac{7}{2}+\frac{\sqrt{65}}{2},y=\frac{7}{2}-\frac{\sqrt{65}}{2}$
C) $y=7+\frac{\sqrt{65}}{2},y=7-\frac{\sqrt{65}}{2}$
D) $y=-\frac{7}{2}+\frac{\sqrt{65}}{2},y=-\frac{7}{2}-\frac{\sqrt{65}}{2}$
E) no solutions

To solve $u^2-7u-4=0$ by completing the square, first move the constant term to the other side: $u^2-7u=4$ . Then, add ${\left(\frac{-7}{2}\right)}^2=\frac{49}{4}$ to both sides to complete the square: $u^2-7u+\frac{49}{4}=4+\frac{49}{4}$ , which simplifies to $u^2-7u+\frac{49}{4}=\frac{65}{4}$ . This gives ${\left(u-\frac{7}{2}\right)}^2=\frac{65}{4}$ . Taking the square root of both sides gives $u-\frac{7}{2}=\pm \frac{\sqrt{65}}{2}$ , leading to $u=\frac{7}{2}\pm \frac{\sqrt{65}}{2}$ .