$\left(\forall x\in\mathbb R\right)\left(x^2+6x+5\geq 0\right)$. I originally was thinking false because when you factor this quadratic equation, it equals zero when $x$ is $-1$ and $x$ is $-5$.
There is another example of this that is in my book and the book says that it is true: $\left(\forall x\in\mathbb R\right)\left(x^2+4x+5\geq 0\right)$
On
The quickest way to check if this is true or not is to find the minimum value of $x^2+6x+5$. Simply take the derivative, set it equal to zero, solve for $x$ and plug that $x$ back into your original equation. If you take the example in your book where $f(x) = x^2+4x+5$ then $f'(x) = 2x+4$ and $f'(x) = 0 \implies x=-2$. Hence, $$f(-2) = (-2)^2+4(-2)+5 = 1 \geq 0 $$ which is why we know the statement $$(\forall x \in \Bbb{R})(x^2+4x+5 \geq 0)$$ is true
On
Think of both of these quadratics as perfect squares that have been shifted upwards (or downwards)
for example $$x^2+6x+5=x^2+6x+9-4=(x+3)^2-4$$ and $$x^2+4x+5=x^2+4x+4+1=(x+2)^2+1$$
The first quadratic has 2 roots (and thus takes on both positive and negative values due to continuity). The second quadratic has no real roots, and being upward opening, is always positive.
Hint: Since the zeros are $-1$ and $-5$, and you know your quadratic opens upward, try a value in between $-1$ and $-5$, like $x = -2$ and see if you get a negative output.