n = \frac{-5 \pm \sqrt5^2 - 4(2)(-150)}2(2) = \frac{-5 \pm \sqrt25 + 1200}4 = \frac{-5 \pm \sqrt1225}4 = \frac-5 \pm 354

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:

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

Understanding the Context

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.

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Image Gallery

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$If \( \frac{1}{2 \cdot 3^{10}}+\frac{2}{2^{2} \cdot 3^{9}}+\frac{3}{2 ...$

Solved: How many? / What type? x^2+5x=2x^2 [algebra]

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Key Insights

Step 1: Identify Coefficients
The 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 Root
Simplify 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:
\[
n = \frac{5 \pm \sqrt{1225}}{4}
\]

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Final Thoughts

Step 3: Compute the Square Root
We 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 Roots
Using 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 Matters
The 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.

n = \frac{-5 \pm \sqrt5^2 - 4(2)(-150)}2(2) = \frac{-5 \pm \sqrt25 + 1200}4 = \frac{-5 \pm \sqrt1225}4 = \frac-5 \pm 354 - Simpleprint

Understanding the Context

Image Gallery

Key Insights

Step 1: Identify Coefficients
The 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 Root
Simplify the discriminant \( b^2 - 4ac \):
\[
(-5)^2 = 25
\]
\[
4 \cdot 2 \cdot (-150) = -1200
\]
\[
b^2 - 4ac = 25 - (-1200) = 25 + 1200 = 1225
\]

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Final Thoughts

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

Step 4: Solve for the Two Roots
Using 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 Matters
The 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 Answer
The solutions to the quadratic equation are:
\[
n = \frac{15}{2} \quad \ ext{and} \quad n = -10
\]

Understanding the Context

Image Gallery

Key Insights

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 \( b^2 - 4ac \):\[(-5)^2 = 25\]\[4 \cdot 2 \cdot (-150) = -1200\]\[b^2 - 4ac = 25 - (-1200) = 25 + 1200 = 1225\]

Continue Reading

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Final Thoughts

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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Step 1: Identify Coefficients
The 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 Root
Simplify 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 Root
We now simplify \( \sqrt{1225} \). Since \( 35^2 = 1225 \),
\[
\sqrt{1225} = 35
\]

Step 4: Solve for the Two Roots
Using 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 Matters
The 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 Answer
The solutions to the quadratic equation are:
\[
n = \frac{15}{2} \quad \ ext{and} \quad n = -10
\]