If $f\left(x\right)=(x^2+3x)(x^2+3x+2)$, then the sum of all real roots of the equation
$$\sqrt{f\left(x\right)}+1=9701$$

The correct option is -3.

To find the sum of all real roots of the given equation, we start by analyzing the function:
$$f\left(x\right)=(x^2+3x)(x^2+3x+2)$$
Let us simplify this expression by introducing a substitution. Let:
$$y=x^2+3x$$
Substituting $y$ into the expression for $f\left(x\right)$, we obtain:
$$f\left(x\right)=y(y+2)=y^2+2y$$

Now, we substitute this back into the given equation:
$$\sqrt{f\left(x\right)}+1=9701$$
Subtracting 1 from both sides of the equation yields:
$$\sqrt{f\left(x\right)}=9700$$
Squaring both sides gives:
$$f\left(x\right)={9700}^2$$
Replacing $f\left(x\right)$ with $y^2+2y$, we get:
$$y^2+2y={9700}^2$$
We can rewrite this equation as a quadratic in terms of $y$:
$$y^2+2y-{9700}^2=0$$

Using the quadratic formula to solve for $y$:
$$y=\frac{-2\±\sqrt{2^2-4\left(1\right)(-{9700}^2)}}{2\left(1\right)}$$
Simplifying the terms:
$$y=\frac{-2\±\sqrt{4+4\·{9700}^2}}{2}$$
Factoring out $4$ from the square root:
$$y=\frac{-2\±2\sqrt{{9700}^2+1}}{2}$$
Dividing by 2 gives:
$$y=-1\±\sqrt{{9700}^2+1}$$

This yields two possible values for $y$:
1) $y_1=-1+\sqrt{{9700}^2+1}$
2) $y_2=-1-\sqrt{{9700}^2+1}$

We now substitute back $y=x^2+3x$ to find the values of $x$.
For a quadratic equation of the form $x^2+3x-y=0$ to have real roots, its discriminant must be non-negative:
$$D=3^2-4\left(1\right)(-y)=9+4y\≥0$$
This means we must have:
$$y\≥-\frac{9}{4}$$

Let us check the discriminant condition for both values of $y$:
• For $y_1=-1+\sqrt{{9700}^2+1}$:
Since $\sqrt{{9700}^2+1}>9700$, we have $y_1>-1+9700=9699$. Clearly, $9699\≥-\frac{9}{4}$. Thus, the quadratic equation $x^2+3x-y_1=0$ has real roots.
• For $y_2=-1-\sqrt{{9700}^2+1}$:
Since $\sqrt{{9700}^2+1}>9700$, we have $y_2<-1-9700=-9701$. This is less than $-\frac{9}{4}$. Therefore, the quadratic equation $x^2+3x-y_2=0$ has no real roots.

Consequently, all real roots of the original equation are the roots of the quadratic equation:
$$x^2+3x-y_1=0$$
By Vieta's formulas, the sum of the roots of a quadratic equation $a{x}^2+bx+c=0$ is given by $-\frac{b}{a}$.
Here, $a=1$ and $b=3$. Thus, the sum of all real roots is:
$$\text{Sum}=-\frac{3}{1}=-3$$