1. **State the problem:** Solve the system of equations given: Top-right: $$n^2 + 4n\left(-\frac{4}{2}\right)^2 = +21\left(\frac{4}{2}\right)^2$$ Top-right (simplified): $$n^2 + 20n = -92$$ Bottom-left: $$6n - 63 = 0$$ 2. **Simplify and solve the bottom-left equation first:** $$6n - 63 = 0$$ Add 63 to both sides: $$6n = 63$$ Divide both sides by 6: $$n = \frac{63}{6}$$ Show cancellation: $$n = \frac{\cancel{63}}{\cancel{6}} = \frac{21}{2} = 10.5$$ 3. **Check if this value satisfies the top-right simplified equation:** $$n^2 + 20n = -92$$ Substitute $$n = 10.5$$: $$10.5^2 + 20 \times 10.5 = ?$$ Calculate: $$110.25 + 210 = 320.25$$ Since $$320.25 \neq -92$$, this value does not satisfy the equation. 4. **Solve the quadratic equation:** $$n^2 + 20n + 92 = 0$$ Use the quadratic formula: $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=20$$, $$c=92$$. Calculate discriminant: $$\Delta = 20^2 - 4 \times 1 \times 92 = 400 - 368 = 32$$ Calculate roots: $$n = \frac{-20 \pm \sqrt{32}}{2} = \frac{-20 \pm 4\sqrt{2}}{2}$$ Simplify: $$n = -10 \pm 2\sqrt{2}$$ 5. **Final answer:** The solutions to the quadratic are: $$n = -10 + 2\sqrt{2}$$ and $$n = -10 - 2\sqrt{2}$$ The linear equation solution $$n=10.5$$ does not satisfy the quadratic equation. Hence, the consistent solutions for $$n$$ are $$-10 \pm 2\sqrt{2}$$.