Answer
$\frac{1}{2}$
Work Step by Step
$k{{x}^{2}}+3x-k=0$
Substitute $x=-2$ to get the value of k as shown below,
$\begin{align}
& k{{\left( -2 \right)}^{2}}+3\left( -2 \right)-k=0 \\
& 4k-6-k=0 \\
& 3k=6 \\
& k=\frac{6}{3}
\end{align}$
Simplify,
$k=2$
Substitute $k=2$ into the provided equation $k{{x}^{2}}+3x-k=0$,
$2{{x}^{2}}+3x-2=0$
Now compare the equation $2{{x}^{2}}+3x-2=0$ with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$,
Here,
$a=2,b=3$ and $c=-2$
Substitute $a=2,b=3$ and $c=-2$ into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$,
$\begin{align}
& x=\frac{-3\pm \sqrt{{{3}^{2}}-4\left( 2 \right)\left( -2 \right)}}{2\left( 2 \right)} \\
& =\frac{-3\pm \sqrt{9+16}}{4} \\
& =\frac{-3\pm \sqrt{25}}{4} \\
& =\frac{-3\pm 5}{4}
\end{align}$
Simplify more,
$\begin{align}
& x=\frac{-3+5}{4} \\
& =\frac{2}{4} \\
& =\frac{1}{2}
\end{align}$
And,
$\begin{align}
& x=\frac{-3-5}{4} \\
& =\frac{-8}{4} \\
& =-2
\end{align}$