Answer
$x=-4$, $x=1$ and $x=-1$ satisfy the given conditions
Work Step by Step
$y=x^{3}+4x^{2}-x+6$ and $y=10$
Substitute $y$ by $10$:
$x^{3}+4x^{2}-x+6=10$
Take all terms to the left side and simplify:
$x^{3}+4x^{2}-x+6-10=0$
$x^{3}+4x^{2}-x-4=0$
Group the first two terms together and the last two terms together:
$(x^{3}+4x^{2})-(x+4)=0$
Take out common factor $x^{2}$ from the first group:
$x^{2}(x+4)-(x+4)=0$
Factor out $x+4$:
$(x+4)(x^{2}-1)=0$
Factor $x^{2}-1$:
$(x+4)(x-1)(x+1)=0$
Set all terms equal to $0$ and solve each individual equation for $x$:
$x+4=0$
$x=-4$
$x-1=0$
$x=1$
$x+1=0$
$x=-1$
$x=-4$, $x=1$ and $x=-1$ satisfy the given conditions