1. **Problem:** Solve the equation $k(2k + 5) = 3$. 2. **Formula and rules:** This is a quadratic equation. Expand and rearrange to standard form $ax^2 + bx + c = 0$. 3. **Work:** $$k(2k + 5) = 3 \implies 2k^2 + 5k = 3$$ Move all terms to one side: $$2k^2 + 5k - 3 = 0$$ 4. **Solve using quadratic formula:** $$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=5$, $c=-3$. Calculate discriminant: $$\Delta = 5^2 - 4 \times 2 \times (-3) = 25 + 24 = 49$$ Calculate roots: $$k = \frac{-5 \pm \sqrt{49}}{2 \times 2} = \frac{-5 \pm 7}{4}$$ Two solutions: $$k_1 = \frac{-5 + 7}{4} = \frac{2}{4} = 0.5$$ $$k_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$ 5. **Answer:** The solutions are $k = 0.5$ and $k = -3$.