Find the sum of all $\left\lfloor x\right\rfloor$ such that $x^2-15\left\lfloor x\right\rfloor+36=0$.

$\text{(A) }15\qquad\text{(B) }26\qquad\text{(C) }45\qquad\text{(D) }49\qquad\text{(E) }75$
1. We start with the given equation:
   \[
   x^2 - 15\left\lfloor x \right\rfloor + 36 = 0
   \]
   Let \( \left\lfloor x \right\rfloor = n \), where \( n \) is an integer. Then the equation becomes:
   \[
   x^2 - 15n + 36 = 0
   \]
   Rearranging, we get:
   \[
   x^2 = 15n - 36
   \]

2. Since \( n \leq x < n+1 \), we have:
   \[
   n^2 \leq 15n - 36 < (n+1)^2
   \]

3. Solving the inequality \( n^2 \leq 15n - 36 \):
   \[
   n^2 - 15n + 36 \leq 0
   \]
   This is a quadratic inequality. To solve it, we find the roots of the corresponding quadratic equation:
   \[
   n^2 - 15n + 36 = 0
   \]
   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -15 \), and \( c = 36 \):
   \[
   n = \frac{15 \pm \sqrt{225 - 144}}{2} = \frac{15 \pm \sqrt{81}}{2} = \frac{15 \pm 9}{2}
   \]
   Thus, the roots are:
   \[
   n = \frac{24}{2} = 12 \quad \text{and} \quad n = \frac{6}{2} = 3
   \]
   Therefore, the inequality \( n^2 - 15n + 36 \leq 0 \) holds for:
   \[
   3 \leq n \leq 12
   \]

4. Next, solving the inequality \( 15n - 36 < (n+1)^2 \):
   \[
   15n - 36 < n^2 + 2n + 1
   \]
   Rearranging, we get:
   \[
   n^2 - 13n + 37 > 0
   \]
   To solve this quadratic inequality, we find the roots of the corresponding quadratic equation:
   \[
   n^2 - 13n + 37 = 0
   \]
   Using the quadratic formula:
   \[
   n = \frac{13 \pm \sqrt{169 - 148}}{2} = \frac{13 \pm \sqrt{21}}{2}
   \]
   The roots are:
   \[
   n = \frac{13 + \sqrt{21}}{2} \quad \text{and} \quad n = \frac{13 - \sqrt{21}}{2}
   \]
   Since these roots are not integers, we analyze the inequality \( n^2 - 13n + 37 > 0 \). The quadratic \( n^2 - 13n + 37 \) is positive outside the interval between the roots. Since \( \frac{13 - \sqrt{21}}{2} \approx 4.21 \) and \( \frac{13 + \sqrt{21}}{2} \approx 8.79 \), the inequality holds for:
   \[
   n < \frac{13 - \sqrt{21}}{2} \quad \text{or} \quad n > \frac{13 + \sqrt{21}}{2}
   \]
   Therefore, the valid integer values of \( n \) are:
   \[
   n \in \{3, 4, 9, 10, 11, 12\}
   \]

5. Summing these values:
   \[
   3 + 4 + 9 + 10 + 11 + 12 = 49
   \]

The final answer is \(\boxed{49}\).