Find the zeros of the function $f(x)=2 x^{2}-8 x+0.9$. Round values to the nearest hundredth (if necessary).

Step 1 ：Given the function \(f(x)=2 x^{2}-8 x+0.9\), we need to find the zeros of the function. The zeros of a function are the x-values for which the function equals zero. In other words, we need to solve the equation \(2x^{2}-8x+0.9=0\) for x.

Step 2 ：This is a quadratic equation, and we can solve it using the quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where a, b, and c are the coefficients of the quadratic equation. In this case, a=2, b=-8, and c=0.9.

Step 3 ：Substitute a, b, and c into the quadratic formula, we get two solutions for x, which are the zeros of the function.

Step 4 ：Calculate the discriminant D = b^2 - 4ac = 56.8.

Step 5 ：Substitute D into the quadratic formula, we get x1 = 3.88 and x2 = 0.12.

Step 6 ：Final Answer: The zeros of the function \(f(x)=2 x^{2}-8 x+0.9\) are \(\boxed{3.88}\) and \(\boxed{0.12}\).