Divide by 2: \( 2n^2 + 5n - 150 = 0 \). Use the quadratic formula: - Dyverse
Solving the Quadratic Equation \(2n^2 + 5n - 150 = 0\) Using the Quadratic Formula
Solving the Quadratic Equation \(2n^2 + 5n - 150 = 0\) Using the Quadratic Formula
When faced with a quadratic equation like \(2n^2 + 5n - 150 = 0\), using the quadratic formula provides a powerful and reliable method to find exact solutions. Whether you're working on math problems, programming algorithms, or scientific modeling, understanding how to apply this formula is essential. In this article, we’ll break down the step-by-step solution to \(2n^2 + 5n - 150 = 0\) using the quadratic formula and explore its application in real-world scenarios.
Understanding the Context
What is the Quadratic Formula?
The quadratic formula solves equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are real numbers and \(a \
e 0\). The formula is:
\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Using this formula, you can find the two roots (real or complex) of any quadratic equation efficiently.
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Key Insights
Step-by-Step Solution to \(2n^2 + 5n - 150 = 0\)
Step 1: Identify coefficients
From the equation \(2n^2 + 5n - 150 = 0\), the coefficients are:
- \(a = 2\)
- \(b = 5\)
- \(c = -150\)
Step 2: Calculate the discriminant
The discriminant, \(D\), tells us about the nature of the roots:
\[
D = b^2 - 4ac
\]
Substitute the values:
\[
D = (5)^2 - 4(2)(-150) = 25 + 1200 = 1225
\]
Since \(D > 0\) and \(D = 1225 = 35^2\), the equation has two distinct real roots.
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Step 3: Apply the quadratic formula
Now substitute into the formula:
\[
n = \frac{-5 \pm \sqrt{1225}}{2 \ imes 2} = \frac{-5 \pm 35}{4}
\]
Step 4: Solve for both roots
-
First root (\(+\) sign):
\[
n_1 = \frac{-5 + 35}{4} = \frac{30}{4} = 7.5
\] -
Second root (\(-\) sign):
\[
n_2 = \frac{-5 - 35}{4} = \frac{-40}{4} = -10
\]
Final Answer
The solutions to the equation \(2n^2 + 5n - 150 = 0\) are:
\[
\boxed{n = 7.5 \quad} \ ext{and} \quad n = -10
\]
