Answer
The solution set is $\left\{ -3+2\sqrt{5},-3-2\sqrt{5} \right\}$.
Work Step by Step
Consider the function.
$f\left( x \right)={{x}^{2}}+6x$
Substitute $x=a $ in $f\left( x \right)={{x}^{2}}+6x$.
Then, $f\left( a \right)={{a}^{2}}+6a$ …… (1)
Given that $f\left( a \right)=11$.
Substitute $f\left( a \right)=11$ in equation (1).
$11={{a}^{2}}+6a$ …… (2)
Subtract $11$ on both sides.
$\begin{align}
& {{a}^{2}}+6a-11=11-11 \\
& {{a}^{2}}+6a-11=0 \\
\end{align}$
The equation ${{a}^{2}}+6a-11=0$ is in quadratic equation form.
Compare the equation ${{a}^{2}}+6a-11=0$ with the general form of the quadratic equation $a{{x}^{2}}+bx+c=0$.
Then, $a=1,b=6$and $c=-11$.
Substitute these corresponding values in the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
& a=\frac{-6\pm \sqrt{{{6}^{2}}-4\left( 1 \right)\left( -11 \right)}}{2\left( 1 \right)} \\
& =\frac{-6\pm \sqrt{36+44}}{2} \\
& =\frac{-6\pm \sqrt{80}}{2}
\end{align}$
Simplify further as:
$\begin{align}
& a=\frac{-6\pm 4\sqrt{5}}{2} \\
& =-3\pm 2\sqrt{5}
\end{align}$
Thus, the solution set for a is $\left\{ -3+2\sqrt{5},-3-2\sqrt{5} \right\}$.
Check:
Substitute $-3+2\sqrt{5}$ for $a$ in equation (2).
$\begin{align}
& 11\overset{?}{\mathop{=}}\,{{\left( -3+2\sqrt{5} \right)}^{2}}+6\left( -3+2\sqrt{5} \right) \\
& 11\overset{?}{\mathop{=}}\,9+20-12\sqrt{5}-18+12\sqrt{5} \\
& 11\overset{?}{\mathop{=}}\,9+20-18 \\
& 11\overset{?}{\mathop{=}}\,11 \\
\end{align}$
It is true.
Substitute $-3-2\sqrt{5}$ for $a$ in equation (2).
$\begin{align}
& 11\overset{?}{\mathop{=}}\,{{\left( -3-2\sqrt{5} \right)}^{2}}+6\left( -3-2\sqrt{5} \right) \\
& 11\overset{?}{\mathop{=}}\,9+20+12\sqrt{5}-18-12\sqrt{5} \\
& 11\overset{?}{\mathop{=}}\,9+20-18 \\
& 11\overset{?}{\mathop{=}}\,11 \\
\end{align}$
It is also true.
