Explanation:
To find the zeros of the function \(h(x) = 2x^2 + 72\), set the function equal to zero and solve for \(x\):
\[2x^2 + 72 = 0\]
First, divide both sides of the equation by 2 to simplify it:
\[x^2 + 36 = 0\]
Now, subtract 36 from both sides:
\[x^2 = -36\]
Since the right side is negative, this equation has no real solutions because the square of any real number is always non-negative. However, if you allow for complex numbers, you can take the square root of both sides:
\[x = \pm 6i\]
So, the zeros of the function \(h(x) = 2x^2 + 72\) are \(x = 6i\) and \(x = -6i\), where \(i\) is the imaginary unit.