Combine like terms: \( 2n^2 + 2n + 1 = 85 \).

["# Combine Like Terms in Quadratic Equations: Solving ( 2n^2 + 2n + 1 = 85 )", "When solving quadratic equations like ( 2n^2 + 2n + 1 = 85 ), one essential step is combining like terms, even if the equation isn’t fully simplified yet. Properly combining terms helps isolate the variable, making it easier to solve for ( n ) using algebraic methods.", "## Understanding the Equation", "Start with the original equation:\n[\n2n^2 + 2n + 1 = 85\n]", "This equation contains a quadratic term (( 2n^2 )), a linear term (( 2n )), and a constant term (1). The constant ( 1 ) is technically a term like ( +1 ), but in this context, it combines seamlessly with the rest. Combining like terms helps reveal the structure needed for further manipulation.", "### Step 1: Move All Terms to One Side", "To combine like terms effectively, move all components to one side of the equation to set it equal to zero — a standard form for solving quadratics:\n[\n2n^2 + 2n + 1 - 85 = 0\n]\nSimplify by combining constants:\n[\n2n^2 + 2n - 84 = 0\n]", "Now, the equation is in standard quadratic form:\n[\n2n^2 + 2n - 84 = 0\n]", "### Step 2: Combine Like Terms (If Needed)", "In this simplified form, observe that all comparable degree terms are already combined:\n- ( 2n^2 ) appears once\n- ( 2n ) appears once\n- Only one constant: ( -84 )", "So, no further combining is necessary — the left side is fully combined, making it ideal for factoring, using the quadratic formula, or completing the square.", "---", "## Why Combine Like Terms Before Solving?", "Combining like terms ensures clarity and avoids errors when manipulating equations. For example:\n- The constant ( +1 ) combined with ( -85 ) to form ( -84 ) prevents confusion during simplification.\n- It reveals the true degree and nature of the polynomial, critical for choosing the correct solving method.", "For ( 2n^2 + 2n + 1 = 85 ):\n- Combine constant terms to get ( 2n^2 + 2n - 84 = 0 )\n- Enables application of the quadratic formula:\n[\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 2, b = 2, c = -84\n]", "---", "## Final Answer", "Using the simplified equation:\n[\n2n^2 + 2n - 84 = 0\n]", "Apply the quadratic formula:\n[\nn = \frac{-2 \pm \sqrt{(2)^2 - 4(2)(-84)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 672}}{4} = \frac{-2 \pm \sqrt{676}}{4}\n]\n[\n\sqrt{676} = 26 \quad \ ext{(since } 26^2 = 676\ ext{)}\n]\n[\nn = \frac{-2 \pm 26}{4}\n]", "So, two solutions:\n[\nn = \frac{24}{4} = 6 \quad \ ext{and} \quad n = \frac{-28}{4} = -7\n]", "---", "### Conclusion", "Combining like terms early — transforming ( 2n^2 + 2n + 1 = 85 ) into ( 2n^2 + 2n - 84 = 0 ) — streamlines the solving process and reduces mistakes. Mastering this technique ensures accurate and efficient solutions in quadratic equations across algebra and advanced mathematics.", "---", "### Key SEO Keywords:\ncombine like terms, solve quadratic equations, quadratic formula tutorial, simplify 2n² + 2n + 1 = 85, algebra step-by-step, solve 2n² + 2n - 84 = 0"]