$$ \gamma $$ and 5 are substracted and added respectively from a positive number. The produce of the rwo news numbers obtoined is 140 Find the number

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Let the positive number be $$ x $$. According to the problem, when $$ \gamma $$ is subtracted from $$ x $$, we get $$ x - \gamma $$, and when 5 is added to $$ x $$, we get $$ x + 5 $$.

The product of these two new numbers is given as:

$$
(x - \gamma)(x + 5) = 140
$$

Expanding this equation, we have:

$$
x^2 + 5x - \gamma x - 5\gamma = 140
$$

Rearranging gives:

$$
x^2 + (5 - \gamma)x - 5\gamma - 140 = 0
$$

This is a quadratic equation in the form $$ ax^2 + bx + c = 0 $$, where:

- $$ a = 1 $$
- $$ b = 5 - \gamma $$
- $$ c = -5\gamma - 140 $$

To find the positive number $$ x $$, we can use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Substituting $$ a $$, $$ b $$, and $$ c $$:

$$
x = \frac{-(5 - \gamma) \pm \sqrt{(5 - \gamma)^2 - 4(1)(-5\gamma - 140)}}{2(1)}
$$

This simplifies to:

$$
x = \frac{\gamma - 5 \pm \sqrt{(5 - \gamma)^2 + 20\gamma + 560}}{2}
$$

To find a specific value for $$ x $$, we need a specific value for $$ \gamma $$. If $$ \gamma $$ is not specified, we cannot find a unique solution for $$ x $$.

If you have a specific value for $$ \gamma $$, please provide it, and we can solve for $$ x $$. Otherwise, the solution will depend on the value of $$ \gamma $$.
The equation is $$ (x - \gamma)(x + 5) = 140 $$. To find the positive number $$ x $$, use the quadratic formula with $$ a = 1 $$, $$ b = 5 - \gamma $$, and $$ c = -5\gamma - 140 $$.

and 5 are substracted and added respectively
from a positive number. The produce of the rwo
news numbers obtoined is 140 Find the number

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