he quadratic equation $$ (1-p) x^{2}-6 x+10=0 $$, where $$ p $$ is a constant, has two different ind the range of values of $$ p $$.

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We start with the quadratic equation
$$
(1-p) x^{2}-6 x+10=0.
$$
For a quadratic equation to have two distinct real roots, two conditions must be met:

1. The quadratic coefficient must be nonzero, i.e., 
   $$
   1-p \neq 0.
   $$
2. The discriminant must be greater than zero. For a quadratic equation in the form
   $$
   ax^2 + bx + c=0,
   $$
   the discriminant is given by 
   $$
   \Delta=b^2-4ac.
   $$

### Step 1. Ensure the Equation is Quadratic

The coefficient of $$x^2$$ is $$(1-p)$$. Thus, to ensure the equation remains quadratic we need
$$
1-p \neq 0 \quad \Rightarrow \quad p \neq 1.
$$

### Step 2. Calculate the Discriminant

Here, the coefficients are:
$$
a=1-p,\quad b=-6,\quad c=10.
$$
The discriminant is then
$$
\Delta=(-6)^2-4(1-p)(10)=36-40(1-p).
$$
Simplify the expression:
$$
36-40(1-p)=36-40+40p=40p-4.
$$

### Step 3. Set the Discriminant Greater Than Zero

For two distinct real roots, we require
$$
40p-4>0.
$$
Solving for $$p$$ gives:
$$
40p>4 \quad \Rightarrow \quad p>\frac{4}{40}=\frac{1}{10}.
$$

### Step 4. Combine Conditions

The two conditions are:
$$
p > \frac{1}{10} \quad \text{and} \quad p \neq 1.
$$

### Final Answer

The quadratic equation has two distinct real roots when
$$
p\in\left(\frac{1}{10},1\right)\cup(1,\infty).
$$
The quadratic equation $$ (1-p) x^{2}-6 x+10=0 $$ has two different real roots when $$ p $$ is greater than $$ \frac{1}{10} $$ and not equal to 1.

he quadratic equation , where is a constant, has two different
ind the range of values of .

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he quadratic equation , where is a constant, has two different ind the range of values of .