-x^2+4x>-5 find solution set. I found the answer as (-1,5) but according to book it must be (1,5)
$-x^2+4x > -5$
$x^2-4x - 5 < 0$
$(x-5)(x+1) < 0$
same thing as the last problem we check signs from $-\infty \to \infty$
$(-\infty, -1) \rightarrow +$
$(-1, 5) \rightarrow -$
$(5, \infty) \rightarrow +$
and so the solution is
$(-1,5)$
and thus you are correct and the book is not!
Or you can solve the inequality more directly by completing the square...
$\displaystyle \begin{align*} -x^2 + 4x &> -5 \\ 0 &> x^2 - 4x - 5 \\ x^2 - 4x - 5 &< 0 \\ x^2 - 4x + (-2) ^2 - (-2) ^2 - 5 &< 0 \\ \left( x - 2 \right) ^2 - 9 &< 0 \\ \left( x - 2 \right) ^2 &< 9 \\ \left| x - 2 \right| &< 3 \\ -3 < x - 2 &< 3 \\ -1 < x &< 5 \end{align*}$