1. **State the problem:** Find the slope of the function $f(x) = \sqrt{5x + 3}$ using the concept of tangent lines and the delta symbol $\Delta$. 2. **Recall the formula for the slope of the tangent line:** The slope of the tangent line at a point $x$ is the derivative $f'(x)$, which can be found using the limit definition: $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ 3. **Apply the formula:** $$f'(x) = \lim_{\Delta x \to 0} \frac{\sqrt{5(x + \Delta x) + 3} - \sqrt{5x + 3}}{\Delta x}$$ 4. **Simplify inside the square roots:** $$f'(x) = \lim_{\Delta x \to 0} \frac{\sqrt{5x + 5\Delta x + 3} - \sqrt{5x + 3}}{\Delta x}$$ 5. **Rationalize the numerator to eliminate the square roots:** Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{5x + 5\Delta x + 3} - \sqrt{5x + 3}}{\Delta x} \times \frac{\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}}{\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}}$$ 6. **Calculate the numerator using difference of squares:** $$\frac{(5x + 5\Delta x + 3) - (5x + 3)}{\Delta x \left(\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}\right)} = \frac{5\Delta x}{\Delta x \left(\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}\right)}$$ 7. **Cancel $\Delta x$ in numerator and denominator:** $$\frac{\cancel{5\Delta x}}{\cancel{\Delta x} \left(\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}\right)} = \frac{5}{\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}}$$ 8. **Take the limit as $\Delta x \to 0$:** $$f'(x) = \lim_{\Delta x \to 0} \frac{5}{\sqrt{5x + 5\Delta x + 3} + \sqrt{5x + 3}} = \frac{5}{\sqrt{5x + 3} + \sqrt{5x + 3}} = \frac{5}{2\sqrt{5x + 3}}$$ 9. **Final answer:** $$\boxed{f'(x) = \frac{5}{2\sqrt{5x + 3}}}$$ This is the slope of the tangent line to the curve $f(x) = \sqrt{5x + 3}$ at any point $x$.