Solving the Quadratic Equation: \frac{n}{2}(4n + 10) = 150 – Step-by-Step Guide

If you've ever encountered an equation like \(\frac{n}{2}(4n + 10) = 150\), you know how powerful algebra can be when solving for unknown variables. This article breaks down how to solve this quadratic equation step by step, showing how to transform, simplify, and apply the quadratic formula to find accurate values of \(n\).

Understanding the Context

Understanding the Equation

We begin with:

\[
\frac{n}{2}(4n + 10) = 150
\]

This equation suggests a proportional relationship multiplied by a linear expression, then set equal to a constant. Solving this will help us uncover the value(s) of \(n\) that satisfy the equation.

Solved: C) (n+1)/n+10 = (n-2)/n+4 [Math]

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Solved: C) (n+1)/n+10 = (n-2)/n+4 [Math]
Solved ∑n=0∞n2+7n+1010 Thes 0/8 ∑n=0∞(1001/(n+2)−1001/(n+3)) | Chegg.com
Solved ∑n=0∞n!n3 ∑n=0∞n2+1n ∑n=0∞10nn! | Chegg.com
Solved 4n+10 6. Let A = [0,1] (-1)" 4n Which of the | Chegg.com
Solved ∑n=130(n2+2)∑n=1020(n3−2n)∑n=010(4i−2i) | Chegg.com

Key Insights

Step 1: Eliminate the fraction

To simplify, multiply both sides of the equation by 2:

\[
2 \cdot \frac{n}{2}(4n + 10) = 2 \cdot 150
\]

\[
n(4n + 10) = 300
\]

Continue Reading

Let \( x \) be the grams of pure salt to be added.
The initial amount of salt is \( 0.02 \times 300 = 6 \) grams.
The final volume of the solution after adding \( x \) grams of salt is \( 300 + x \) mL.

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Final Thoughts

Now we expand the left-hand side.

Step 2: Expand the quadratic expression

Distribute \(n\) across the parentheses:

\[
n \cdot 4n + n \cdot 10 = 4n^2 + 10n
\]

So the equation becomes:

\[
4n^2 + 10n = 300
\]

Step 3: Bring all terms to one side

To form a standard quadratic equation, subtract 300 from both sides: