Roots: \( x = rac{4 \pm \sqrt{16 - 12}}{2} = rac{4 \pm 2}{2} = 3, 1 \)

["Roots of the Quadratic Equation: Solving ( x = \frac{4 \pm \sqrt{16 - 12}}{2} = 3 ) and ( x = 1 )", "Solving quadratic equations is a fundamental skill in algebra, and understanding how to find the roots of expressions like ( x = \frac{4 \pm \sqrt{16 - 12}}{2} ) is essential for students and learners alike. This breakdown explains step-by-step how to calculate the roots and avoids common pitfalls using clear and accessible language.", "---", "### Understanding the Quadratic Formula\nThe general form of a quadratic equation is:\n[\nax^2 + bx + c = 0\n]\nTo find the roots, we use the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nIn this expression, the discriminant ( b^2 - 4ac ) determines the nature of the solutions—whether they are real and distinct, real and repeated, or complex.", "---", "### Step-by-Step Solution\nLet’s analyze the given equation:\n[\nx = \frac{4 \pm \sqrt{16 - 12}}{2}\n]\nHere, ( a = 1 ), ( b = 0 ), ( c = -12 ) (from expanding ( x = \frac{4 \pm \sqrt{16}}{2} ), noting ( \sqrt{16 - (4 \cdot 1 \cdot -12)} = \sqrt{16 + 48} = \sqrt{64} = 8 ); but wait — this step-by-step matches the provided simplification).", "#### 1. Simplify the Discriminant\nStart with the discriminant:\n[\nb^2 - 4ac = 0^2 - 4(1)(-12) = 0 + 48 = 48\n]\nBut the original simplification shows ( \sqrt{16 - 12} = \sqrt{4} = 2 ), which conflicts—let’s clarify:", "Wait: the expression ( \sqrt{16 - 12} ) appears simplified, implying ( b^2 = 16 ) and ( 4ac = 12 ). Since ( b = 0 ), this does not align directly—hence, the full expansion is key.", "From earlier:\n[\nx = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2}\n]\nNow compute both solutions:\n[\nx = \frac{4 + 2}{2} = \frac{6}{2} = 3, \quad x = \frac{4 - 2}{2} = \frac{2}{2} = 1\n]\nThus, the roots are ( x = 3 ) and ( x = 1 ).", "---", "### Why the Discriminant Matters\nThe discriminant ( D = 16 - 48 = -32 )—wait, there’s a discrepancy. Let's correct:", "Actually, ( 16 - 12 = 4 ), meaning ( \sqrt{4} = 2 ), not 48. Therefore, the correct discriminant is ( \sqrt{4} = 2 ), not involving 48. So:\n[\nx = \frac{4 \pm 2}{2} \Rightarrow 3 \ ext{ and } 1\n]\nThis confirms the result.", "---", "### Key Takeaways\n- Always verify the discriminant’s sign: positive → two real roots, zero → one real root, negative → complex roots.\n- When ( b = 0 ), the quadratic simplifies using only coefficients ( a ), ( c ).\n- Using the quadratic formula strictly avoids sign errors or miscalculations.\n- Simplifying step-by-step improves clarity and reduces mistakes.", "---", "### Final Answer\nThe solutions to the equation are:\n[\n\boxed{x = 3} \quad \ ext{and} \quad \boxed{x = 1}\n]\nThese roots arise from correctly applying the quadratic formula with attention to signs and simplifications—critical for mastering algebra.", "---", "Boost Your Learning:** Practice solving similar quadratic equations and explore how changing coefficients alters the discriminant and root types. Mastering these techniques enhances problem-solving skills essential in mathematics and related fields."]