Simplifiez : \( 300v + 6000 + 300v = 7v^2 + 140v \).

["Simplifying the Equation: Solving ( 300v + 6000 + 300v = 7v^2 + 140v )", "Solving quadratic equations can seem daunting at first, but with careful simplification and step-by-step handling, even complex expressions become manageable. Today, we break down and simplify the equation:\n[\n300v + 6000 + 300v = 7v^2 + 140v\n]", "---", "### Step 1: Combine Like Terms on the Left Side", "On the left-hand side, combine the like terms involving ( v ):\n[\n300v + 300v = 600v\n]\nSo the equation becomes:\n[\n600v + 6000 = 7v^2 + 140v\n]", "---", "### Step 2: Bring All Terms to One Side to Form a Standard Quadratic", "Subtract ( 600v + 6000 ) from both sides to move everything to the right:\n[\n0 = 7v^2 + 140v - 600v - 6000\n]", "Now simplify:\n[\n0 = 7v^2 - 460v - 6000\n]", "---", "### Step 3: Rewrite in Standard Quadratic Form", "The equation is now in standard form:\n[\n7v^2 - 460v - 6000 = 0\n]", "This is a quadratic equation of the form ( av^2 + bv + c = 0 ), where:\n- ( a = 7 )\n- ( b = -460 )\n- ( c = -6000 )", "---", "### Step 4: Simplify the Equation (Optional)", "Since the coefficients have no common factor, the equation is already simplified. However, dividing through by 7 can reduce decimal coefficients in further calculation:\n[\nv^2 - \frac{460}{7}v - \frac{6000}{7} = 0\n]\nBut for solving, keeping integer coefficients is typically preferred.", "---", "### Step 5: Solve Using the Quadratic Formula", "Use the quadratic formula:\n[\nv = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nSubstitute ( a = 7 ), ( b = -460 ), ( c = -6000 ):\n[\nv = \frac{460 \pm \sqrt{(-460)^2 - 4(7)(-6000)}}{2(7)}\n]\nCalculate discriminant:\n[\n(-460)^2 = 211600,\quad 4 \cdot 7 \cdot 6000 = 168000\n]\n[\n\ ext{Discriminant} = 211600 + 168000 = 379600\n]\nNow compute:\n[\nv = \frac{460 \pm \sqrt{379600}}{14}\n]", "Now simplify ( \sqrt{379600} ).\nNote:\n[\n\sqrt{379600} = \sqrt{400 \ imes 949} = 20\sqrt{949}\n]\nSince 949 is not a perfect square and has no square factors, we leave it as is or approximate:\n[\n\sqrt{379600} \approx 616\n]\n(Check: ( 616^2 = 379456 ), close; actual value ≈ 616.06)", "Now compute approximate values:\n[\nv = \frac{460 \pm 616.06}{14}\n]", "Calculate both roots:\nFirst root (positive sign):\n[\nv = \frac{460 + 616.06}{14} = \frac{1076.06}{14} \approx 76.72\n]\nSecond root (negative sign):\n[\nv = \frac{460 - 616.06}{14} = \frac{-156.06}{14} \approx -11.15\n]", "---", "### Summary: Simplified Eventual Roots", "The simplified exact form comes from:\n[\nv = \frac{460 \pm \sqrt{379600}}{14}\n]\nBut numerically, the solutions are approximately:\n[\nv \approx 76.72 \quad \ ext{and} \quad v \approx -11.15\n]", "---", "### Why Simplify Equations Like This?", "Simplifying complex expressions helps in:\n- Solving quadratic equations accurately\n- Minimizing computational error\n- Understanding symmetry and structure in equations\n- Applying mathematical tools more effectively", "---", "### Practical Tip", "When solving equations like ( Av^2 + Bv + C = 0 ), always:\n1. Combine like terms\n2. Move all terms to one side to get standard form\n3. Apply the quadratic formula cleanly\n4. Simplify radicals or coefficients when possible", "---", "Key takeaway: Simplifying equations step-by-step makes difficult problems approachable and reveals clear solutions waiting beneath the surface.", "For more math simplification guides, check out algebra basics and quadratic formula tutorials!", "---", "Keywords: Simplify ( 300v + 6000 + 300v = 7v^2 + 140v ), solve quadratic equation, step-by-step solution, quadratic formula application, simplify ( 7v^2 - 460v - 6000 = 0 )", "---", "Meta Description: Learn how to simplify and solve ( 300v + 6000 + 300v = 7v^2 + 140v ) using step-by-step algebra and approximate numerical solutions. Perfect for students and self-learners!"]