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.
 A: I'm thinking that with the algebra/precalculus tag, the equation you're asking about is $$s^2+\frac {7}{4} s = \frac {1}{2}$$
Completing the square is correct; when you have
$$s^2 + \frac {7}{4}s + \frac {49}{64} = \frac {1}{2} + \frac {49}{64}$$ you can factor the left hand side; then you'll have $$\left(s + \frac {7}{8} \right)^2=\frac {81}{64}$$ Take the square roots of both sides, subtract $\dfrac {7}{8}$, and you're finished.  (You should get $\dfrac {1}{4}$ and $-2$ as your final solutions.)
Note: If you had intended to solve the equation $$s^2+\frac {7}{4s} = \frac {1}{2}$$ you can multiply by $4s$ and then you'll have an cubic equation $$4s^3-2s+7 = 0$$ The solution to this is not nice; the best way to solve this is by numerical methods.
A: 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.
