1. **State the problem:** Solve algebraically the equation $$\sqrt[2]{19a + 6} - 2a = 3$$. 2. **Rewrite the problem:** Since the index is 2, the radical is a square root. So the equation is $$\sqrt{19a + 6} - 2a = 3$$. 3. **Isolate the square root:** Add $2a$ to both sides: $$\sqrt{19a + 6} = 3 + 2a$$ 4. **Square both sides to eliminate the square root:** $$\left(\sqrt{19a + 6}\right)^2 = (3 + 2a)^2$$ $$19a + 6 = (3 + 2a)^2$$ 5. **Expand the right side:** $$(3 + 2a)^2 = 3^2 + 2 \times 3 \times 2a + (2a)^2 = 9 + 12a + 4a^2$$ 6. **Write the equation:** $$19a + 6 = 9 + 12a + 4a^2$$ 7. **Bring all terms to one side:** $$0 = 9 + 12a + 4a^2 - 19a - 6$$ $$0 = 4a^2 + (12a - 19a) + (9 - 6)$$ $$0 = 4a^2 - 7a + 3$$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=4$, $b=-7$, $c=3$. Calculate the discriminant: $$\Delta = (-7)^2 - 4 \times 4 \times 3 = 49 - 48 = 1$$ Calculate the roots: $$a = \frac{-(-7) \pm \sqrt{1}}{2 \times 4} = \frac{7 \pm 1}{8}$$ So, $$a_1 = \frac{7 + 1}{8} = \frac{8}{8} = 1$$ $$a_2 = \frac{7 - 1}{8} = \frac{6}{8} = \frac{3}{4}$$ 9. **Check for extraneous solutions:** Substitute $a=1$ into the original equation: $$\sqrt{19(1) + 6} - 2(1) = \sqrt{25} - 2 = 5 - 2 = 3$$ (True) Substitute $a=\frac{3}{4}$: $$\sqrt{19 \times \frac{3}{4} + 6} - 2 \times \frac{3}{4} = \sqrt{\frac{57}{4} + 6} - \frac{3}{2} = \sqrt{\frac{57}{4} + \frac{24}{4}} - \frac{3}{2} = \sqrt{\frac{81}{4}} - \frac{3}{2} = \frac{9}{2} - \frac{3}{2} = 3$$ (True) Both solutions are valid. **Final answer:** $$a = 1 \text{ or } a = \frac{3}{4}$$