The roots are both real when the discriminant of the quadratic is positive:
\(a^2-4(a+15)=a^2-4a-60=(a-10)(a+6)>0\)
When \(a<-6\), say \(a=-7\), you have \((-7-10)(-7+6)=17>0\).
When \(-6, say \(a=0\), you have \((0-10)(0+6)=-60<0\).
When \(a>10\), say \(a=11\), you have \((11-10)(11+6)=17>0\).
So the quadratic will have two real roots whenever \(a<-6\) or \(a>10\). The probability of this occurring is
\(\mathbb P((a<-6)\lor(a>10))=\mathbb P(a<-6)+\mathbb P(a>10)\)
The density function for this random variable is
\(f_X(x)=\begin{cases}\dfrac1{11-(-15)}=\dfrac1{26}&\text{for }-15\le x\le11\\\\0&\text{otherwise}\end{cases}\)
so
\(\mathbb P(a<-6)+\mathbb P(a>10)=\displaystyle\int_{-15}^{-6}\dfrac{\mathrm dx}{26}+\int_{10}^{11}\dfrac{\mathrm dx}{26}=\dfrac{(-6-(-15))+(11-10)}{26}=\dfrac5{13}\)
