Answer
$\frac{-9\pm \sqrt{85}}{2}$.
Work Step by Step
${{x}^{2}}+9x=1$.
Now, subtract $1$ from both sides:
$\begin{align}
& {{x}^{2}}+9x=1 \\
& {{x}^{2}}+9x-1=1-1 \\
& {{x}^{2}}+9x-1=0
\end{align}$
Now compare the equation ${{x}^{2}}+9x=1$ with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$,
Therefore,
$a=1,b=9\text{ and }c=-1$
Substitute $a=1,b=9\text{ and }c=-1$into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$,
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-\left( 9 \right)\pm \sqrt{{{\left( 9 \right)}^{2}}-4\cdot 1\cdot \left( -1 \right)}}{2\cdot 1} \\
& =\frac{-9\pm \sqrt{81+4}}{2} \\
& =\frac{-9\pm \sqrt{85}}{2}
\end{align}$
Simplify further as shown below,
$x=\frac{-9+\sqrt{85}}{2}$
And,
$x=\frac{-9-\sqrt{85}}{2}$
Therefore, the solution of the provided expression ${{x}^{2}}+9x=1$ is $\frac{-9\pm \sqrt{85}}{2}$.
