Question

find all real solutions for the equation $n^2 + 18n + 72 = 0$. separate multiple solutions with commas if necessary.
$n =$

Explanation:

Step1: Identify the quadratic formula

For a quadratic equation \(ax^2 + bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 18\), \(c = 72\).

Step2: Calculate the discriminant

The discriminant \(D=b^2 - 4ac\). Substitute the values: \(D = 18^2-4\times1\times72=324 - 288 = 36\).

Step3: Find the solutions

Substitute \(a\), \(b\), and \(D\) into the quadratic formula:
\(n=\frac{-18\pm\sqrt{36}}{2\times1}=\frac{-18\pm6}{2}\)
For the plus sign: \(n=\frac{-18 + 6}{2}=\frac{-12}{2}=-6\)
For the minus sign: \(n=\frac{-18 - 6}{2}=\frac{-24}{2}=-12\)

Answer:

-6, -12