Answer
$x=-3-\sqrt{14}
\text{ and }
x=-3+\sqrt{14}
$
Work Step by Step
Using the properties of equality, the given equation, $
6x-5=-x^2
,$ is equivalent to
\begin{align*}
x^2+6x-5=0
\end{align*}
In the equation above, $a=
1
,$ $b=
6
,$ and $c=
-5
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-6\pm\sqrt{6^2-4(1)(-5)}}{2(1)}
\\\\&=
\dfrac{-6\pm\sqrt{36+20}}{2}
\\\\&=
\dfrac{-6\pm\sqrt{56}}{2}
\\\\&=
\dfrac{-6\pm\sqrt{4\cdot14}}{2}
\\\\&=
\dfrac{-6\pm2\sqrt{14}}{2}
\\\\&=
\dfrac{-\cancel6^3\pm\cancel2^1\sqrt{14}}{\cancel2^1}
&\text{ (divide by $2$)}
\\\\&=
-3\pm\sqrt{14}
\end{align*}
Hence, the solutions are $
x=-3-\sqrt{14}
\text{ and }
x=-3+\sqrt{14}
.$
