n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-150)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 1200}}{4} = \frac{-5 \pm \sqrt{1225}}{4} = \frac{-5 \pm 35}{4}

February 17, 2026

$n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-150)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 1200}}{4} = \frac{-5 \pm \sqrt{1225}}{4} = \frac{-5 \pm 35}{4}$

Solving Quadratic Equations: A Step-by-Step Guide Using the Quadratic Formula

Mastering quadratic equations is essential in algebra, and one of the most powerful tools for solving them is the quadratic formula:

\[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In this article, we walk through a practical example using the equation:

\[n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-150)}}{2(2)}\]

This equation models real-world problems involving area, projectile motion, or optimization—common in science, engineering, and economics. Let’s break down the step-by-step solution and explain key concepts to strengthen your understanding.

Step 1: Identify CoefficientsThe general form of a quadratic equation is:\[an^2 + bn + c = 0\]From our equation:- $ a = 2 $- $ b = -5 $- $ c = -150 $

Plugging these into the quadratic formula gives:\[n = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-150)}}{2(2)}\]

Step 2: Simplify Inside the Square RootSimplify the discriminant $ b^2 - 4ac $:\[(-5)^2 = 25\]\[4 \cdot 2 \cdot (-150) = -1200\]\[b^2 - 4ac = 25 - (-1200) = 25 + 1200 = 1225\]

So far, the equation reads:\[n = \frac{5 \pm \sqrt{1225}}{4}\]

Step 3: Compute the Square RootWe now simplify $ \sqrt{1225} $. Since $ 35^2 = 1225 $,\[\sqrt{1225} = 35\]

Now the expression becomes:\[n = \frac{-5 \pm 35}{4}\](Note: Because $ -b = -(-5) = 5 $, the numerator is $ 5 \pm 35 $.)

Step 4: Solve for the Two RootsUsing the ± property, calculate both solutions:1. $ n_1 = \frac{-5 + 35}{4} = \frac{30}{4} = \frac{15}{2} = 7.5 $2. $ n_2 = \frac{-5 - 35}{4} = \frac{-40}{4} = -10 $

Why This Method MattersThe quadratic formula provides exact solutions—even when the discriminant yields a perfect square like 1225. This eliminates errors common with approximation methods and allows precise modeling of physical or financial systems.

Applications include maximizing profit, determining roots of motion paths, or designing optimal structures across STEM fields.

Final AnswerThe solutions to the quadratic equation are:\[n = \frac{15}{2} \quad \ ext{and} \quad n = -10\]

Understanding this workflow equips you to solve quadratic equations confidently—critical whether you’re a student, teacher, or professional. Mastering this formula opens doors to deeper mathematical and analytical skills!

Keywords: quadratic formula, solving quadratics, discriminant, roots of equations, algebra tutorial, 5^2 - 4ac, real-world math applications, educational math guide

n = \frac{-5 \pm \sqrt{5^2 - 4(2)(-150)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 1200}}{4} = \frac{-5 \pm \sqrt{1225}}{4} = \frac{-5 \pm 35}{4}

Step 1: Identify CoefficientsThe general form of a quadratic equation is:\[an^2 + bn + c = 0\]From our equation:- \( a = 2 \)- \( b = -5 \)- \( c = -150 \)

Step 2: Simplify Inside the Square RootSimplify the discriminant \( b^2 - 4ac \):\[(-5)^2 = 25\]\[4 \cdot 2 \cdot (-150) = -1200\]\[b^2 - 4ac = 25 - (-1200) = 25 + 1200 = 1225\]

Step 3: Compute the Square RootWe now simplify \( \sqrt{1225} \). Since \( 35^2 = 1225 \),\[\sqrt{1225} = 35\]

Step 4: Solve for the Two RootsUsing the ± property, calculate both solutions:1. \( n_1 = \frac{-5 + 35}{4} = \frac{30}{4} = \frac{15}{2} = 7.5 \)2. \( n_2 = \frac{-5 - 35}{4} = \frac{-40}{4} = -10 \)

Why This Method MattersThe quadratic formula provides exact solutions—even when the discriminant yields a perfect square like 1225. This eliminates errors common with approximation methods and allows precise modeling of physical or financial systems.

Final AnswerThe solutions to the quadratic equation are:\[n = \frac{15}{2} \quad \ ext{and} \quad n = -10\]

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