Vse \( b^{2}-4 a c \) Oetermme the real number of solutions of the equatun \( -x^{2}+4 x-4=0 \)

To determine the number of real solutions to the equation \(-x^{2} + 4x - 4 = 0\), we start by identifying the coefficients \(a\), \(b\), and \(c\) in the standard form \(ax^{2} + bx + c = 0\). Here, \(a = -1\), \(b = 4\), and \(c = -4\). The discriminant is calculated using the formula \(b^{2} - 4ac\). Plugging in the values, we have \(4^{2} - 4(-1)(-4) = 16 - 16 = 0\). Since the discriminant is zero, there is exactly one real solution to the equation. Now, this single solution indicates that the quadratic function touches the x-axis at a single point, known as a double root. You can also find this root using the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). In this case, it simplifies nicely to \(x = \frac{-4 + 0}{-2} = 2\). So, the parabola reaches its vertex at this point and does not cross the x-axis.