$$ y=-2 x^{2}+10 x-8 $$ and $$ y=x^{2}-2 x-8 $$

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Sure, let's analyze the two quadratic equations you've provided:

1. $$ y = -2x^{2} + 10x - 8 $$
2. $$ y = x^{2} - 2x - 8 $$

Assuming you want to find the points where these two curves intersect, we'll set the equations equal to each other and solve for $$ x $$.

### Step 1: Set the Equations Equal to Each Other
$$
-2x^{2} + 10x - 8 = x^{2} - 2x - 8
$$

### Step 2: Move All Terms to One Side to Form a Quadratic Equation
Subtract $$ x^{2} - 2x - 8 $$ from both sides:
$$
-2x^{2} + 10x - 8 - (x^{2} - 2x - 8) = 0
$$
$$
-2x^{2} + 10x - 8 - x^{2} + 2x + 8 = 0
$$
$$
-3x^{2} + 12x = 0
$$

### Step 3: Simplify the Equation
$$
-3x^{2} + 12x = 0
$$
Factor out the common term:
$$
-3x(x - 4) = 0
$$

### Step 4: Solve for $$ x $$
Set each factor equal to zero:
1. $$ -3x = 0 $$ ⟹ $$ x = 0 $$
2. $$ x - 4 = 0 $$ ⟹ $$ x = 4 $$

### Step 5: Find the Corresponding $$ y $$-Values
Plug each $$ x $$-value back into either of the original equations to find $$ y $$.

**For $$ x = 0 $$:**
Using $$ y = -2x^{2} + 10x - 8 $$:
$$
y = -2(0)^{2} + 10(0) - 8 = -8
$$
So, one intersection point is $$ (0, -8) $$.

**For $$ x = 4 $$:**
Using $$ y = -2x^{2} + 10x - 8 $$:
$$
y = -2(4)^{2} + 10(4) - 8 = -2(16) + 40 - 8 = -32 + 40 - 8 = 0
$$
So, the other intersection point is $$ (4, 0) $$.

### Conclusion
The two quadratic equations intersect at the following points:
- **Point 1:** $$ (0, -8) $$
- **Point 2:** $$ (4, 0) $$

If you need further analysis, such as graphing these equations or exploring their properties, feel free to ask!
The two equations intersect at the points $$ (0, -8) $$ and $$ (4, 0) $$.

\( y=-2 x^{2}+10 x-8 \) and \( y=x^{2}-2 x-8 \)

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