Question

x$x^{2} -4x-5 \leq 0$

Answer 1

$x^2-4x-5 \leq 0 \\ x^2+x-5x-5 \leq 0 \\ x(x+1)-5(x+1) \leq 0 \\ (x-5)(x+1) \leq 0$

a>0 ⇒ x∈[-1;5]