Hence solve the equation $$ -x^{2}+4 x+1=0 $$, giving your answers correct to two decimal places.

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To solve the quadratic equation $$-x^{2} + 4x + 1 = 0$$, follow these steps:

1. **Rewrite the Equation:**
   $$
   -x^{2} + 4x + 1 = 0 \quad \Rightarrow \quad x^{2} - 4x - 1 = 0
   $$
   (Multiplying both sides by $$-1$$ for simplicity.)

2. **Use the Quadratic Formula:**
   The quadratic formula is:
   $$
   x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
   $$
   For the equation $$x^{2} - 4x - 1 = 0$$, the coefficients are:
   $$
   a = 1, \quad b = -4, \quad c = -1
   $$
   
3. **Calculate the Discriminant ($$b^{2} - 4ac$$):**
   $$
   \Delta = (-4)^2 - 4(1)(-1) = 16 + 4 = 20
   $$
   
4. **Find the Roots:**
   $$
   x = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}
   $$
   
5. **Calculate the Decimal Values:**
   $$
   \sqrt{5} \approx 2.23607
   $$
   
   So,
   $$
   x_1 = 2 + 2.23607 \approx 4.23607 \quad \Rightarrow \quad 4.24 \ (\text{to two decimal places})
   $$
   $$
   x_2 = 2 - 2.23607 \approx -0.23607 \quad \Rightarrow \quad -0.24 \ (\text{to two decimal places})
   $$

**Final Answer:**
$$
x = 4.24 \quad \text{and} \quad x = -0.24
$$
The solutions to the equation $$-x^{2} + 4x + 1 = 0$$ are $$x = 4.24$$ and $$x = -0.24$$.

Hence solve the equation , giving your answers correct to two
decimal places.

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Hence solve the equation , giving your answers correct to two decimal places.