[tex]x^2+6x > -5\ \ | \ add\ 5\ to\ both\ sides \\\\x^2+6x+5>0 \\ \\(x^2+x)+(5x+5)>0 \\\\ x(x+1)+5(x+1)>0\\\\(x+1)(x+5)>0 \\ \\first\ we \ solve: \\(x+1)(x+5)=0\\x+1=0 \ \ or \ \ x+5=0\\x=-1 \ \ or \ \ x=-5[/tex]
a > 0 open for up .
So we know that the curve is u-shaped and that it crosses the x-axis at x=-1 and x=-5.
The curve is more than zero : -5< x < -1