Getting square root of negative in completing the square problem I try to solve the equation $f(x) = 7x - 11 - 2x^2 = 0$ for $x$, but run into troubles. I've gone through it over and over again as well as similar problems, but can't find what I'm doing wrong.
$$f(x) = 7x - 11 - 2x^2 = 0$$
$$\iff x^2 - \frac{7}{2}x + \frac{11}{2} = 0 $$
$$\iff \left(x + \frac{7}{4}\right)^2 = \left(\frac{7}{4}\right)^2 - \frac{11}{2}$$
$$\iff x + \frac{7}{4} = \pm \sqrt{\left(\frac{7}{4}\right)^2 - \frac{11}{2}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{49}{16} - \frac{88}{16}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{-39}{16}}$$
I should be able to continue but I'm stuck (seeing as it's a negative number). What am I doing wrong?

You did everything fine but your quadratic equation has no real solutions, which you could have found out way more easily had you first calculated the equation's discriminant:
$$\Delta:=b^2-4ac=7^2-4(-2)(-11)=49-88=-39<0$$
