How do you finish the equations for $S^2 + \frac{7}{4}S= \frac{1}{2}$? Complete the Square:
$$S^2 + \frac{7}{4}S = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S \_\_\_\_\_\_\_ = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S + \frac{49}{64} = \frac{81}{64}$$
THIS is where I get stuck. I know that the Square Root of $\frac{81}{64}$ is $\frac{9}{8}$, but how do I find the Square Root of $S^2 + \frac{7}{4}S + \frac{49}{64}$?
Quadratic Equation:
$$A=1,\ \ B=\frac{7}{4},\ \ C= \frac{-1}{2}$$
$$\frac{\frac{-7}{4} \pm 
\sqrt{\frac{49}{16}}  + 2}{2}$$
THIS is where I get stuck because I don't know what to do with the 2 in the discriminant.

When you complete the square, assuming the coefficient on the quadratic term is $1$, you take the coefficient on the linear term and divide by two and then square it.  That gives you two numbers,
$b/2$ and $b^2/4$.  Don't throw away the $b/2$.  See where it comes in:
$$x^2+bx = c$$
$$x^2+bx+\frac{b^2}{4} = c+\frac{b^2}{4}$$
$$\left(x+\frac{b}{2}\right)^2 = c+\frac{b^2}{4}.$$
So you took $7/4$, divided by two to get $7/8$.  Keep track of the $7/8$.  Then squared it to get $49/64$.  So the left side factors as $(s+7/8)^2$.   If you multiply it out, you should see why this is always true.
