Extracting Square Roots: Solving the equation

First, we should notice that it is a quadratic equation because there is a quadratic term (meaning raised to the second power).

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side.

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula.

Let's recall it first:

The rule states that the roots of an equation in the form$ax^2+bx+c=0$ are $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

$$This formula is called: "The Quadratic Formula"

Let's now solve the problem:

$4x^2+1=0$

First, let's identify the coefficients of the terms:

$\begin{cases}a=4 \\ b=0 \\ c=1\end{cases}$

Note that in the given equation there is no first-power term, so from comparing to the general form:

$ax^2+bx+c=0$

we can conclude that the coefficient $b$ (which is the coefficient of the first-power term $x$ in the general form) is 0.

Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{0\pm\sqrt{0^2-4\cdot4\cdot1}}{2\cdot4}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{\pm\sqrt{-16}}{8}$

We now have a negative expression under the root and since we cannot extract a real root from a negative number, this equation has no real solutions.

In other words, there is no real value of $x$ that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.