It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $ f(x) := x^4-1102x^3-2015$, we are done.
Observe, $f(0) < 0$ and $f(-2) > 0 $, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.
Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$
Descartes's rule of sign will not give you the exact number as you have shown in your question. Here are some other methods that can be applied to a polynomial $p(x)$.
1) Check whether there are multiple (real or complex) roots. A root $a$ of $p(x)$ is a multiple root $a$ if and only if $a$ is a common root of $p(x), p'(x)$. To see this, write $p(x) = (x-a)^nq(x)$ with $n \ge 1$ and $q(a) \ne 0$. Then $p'(x) = n(x-a)^{n-1}q(x) + (x-a)^nq'(x)$. If $n > 1$, then $p'(a) = 0$ and if $n = 1$ then $p'(a) \ne 0$. We conclude that $p(x)$ has a multiple root if and only if the gcd of $p(x)$ and $p'(x)$ (which can easily be computed by the Euclidean algorithm) has degree $\ge 1$. In fact, if $p(x)$ has a root of $a$ of order $n > 1$, then $(x-a)^{n-1}$ divides $p(x), p'(x)$. Conversely, assume that the gcd $h(x)$ has degree $\ge 1$. Then $h(x)$ has a root $a$ and we see that $p(a) = p'(a) = 0$.
Applying this to your polynomial $p(x) = x^5+5x^4-\frac{65}{3}x^3-70x^2 +300x +297$ shows that no multiple roots exist. Hence it has $1$ or $3$ aor $5$ real roots of order $1$ (since non-real roots occur in complex conjugate pairs).
2) Consider $p'(x), p''(x),p'''(x)$ to find extrema and inflection points. This allows you to determine roots of $p(x)$ lying between these points.
Doing this with your polymomial yields $p'(x) = 5x^4+20x^3-65x^2-140x+300 = 5(x^4+4x^3-13x^2-28x+60)$, $p''(x) = 20x^3+60x^2-130x-140 = 10(2x^3+6x^2-13x-14)$, $p'''(x) = 60x^2+120x-130 = 10(6x^2-12x-13)$. It is easy to verify that $p'(x)$ has the roots $-5,-3,2$, where $2$ has order $2$ (try the divisors of $60$). We have $p''(-5) = -21 < 0$, $p''(-3) = 53 > 0$, $p''(2) = 0$, $p'''(2) = 250 \ne 0$. Hence $p(x)$ has a local maxinum at $-5$, a local minimum at $-3$ and an inflection point at $2$. Hence $p(x)$ is increasing on $(-\infty,-5]$, decreasing on $[-5,-3]$ and increasing on $[-3,\infty)$. We have $p(-5) = -\frac{734}{3}$, $p(-3) = -486$. Hence $p(x)$ does not have a root in $(-\infty,-3]$ and a single root of order $1$ n $[-3,\infty)$.
Best Answer
Given that the roots of $ax^2+bx+c$ are both positive, and real, let the roots be $x=\alpha, \beta>0$ which follows from the condition. Note that the quadratic with the absolute value can be changed into $$ax^2+b|x|+c=0 \iff a|x|^2+b|x|+c=0$$ So we have$$|x|=\alpha, \beta$$ is a root. So there are four roots, $\alpha$, $-\alpha$, $\beta$, $-\beta$.