We are given the radical equation:
[tex]\[ 3 \sqrt[4]{x^2 - x + 4} + 6 = 0 \][/tex]
First, let's isolate the radical term by subtracting 6 from both sides of the equation:
[tex]\[ 3 \sqrt[4]{x^2 - x + 4} = -6 \][/tex]
Next, we divide both sides of the equation by 3:
[tex]\[ \sqrt[4]{x^2 - x + 4} = -2 \][/tex]
Now, let's consider the properties of the fourth root. The fourth root of any real number is always non-negative because raising a real number to the fourth power will always yield a non-negative result. This means:
[tex]\[ \sqrt[4]{x^2 - x + 4} \geq 0 \][/tex]
Since the equation [tex]\(\sqrt[4]{x^2 - x + 4} = -2\)[/tex] indicates that the fourth root must equal a negative number (-2), we see a contradiction. Because a non-negative number (the left-hand side) cannot equal a negative number (the right-hand side), the original equation
[tex]\[ 3 \sqrt[4]{x^2 - x + 4} + 6 = 0 \][/tex]
has no real solutions.
Therefore, the number of real solutions to the equation is:
[tex]\[ \boxed{1} \][/tex]
Option \#1: The equation has 0 real solutions.
