solving the equation \(x^2 + \sqrt{p} x+q=0 yields \Delta= p-4q\)

the equation has at least one real solution if Delta>=0 meaning p>=4q so q has to be from -b to p/4 thus the probability is
\( \frac{b+p/4}{2b} =1/2 + \frac{p}{4b}   \)

thus \( dP = \frac{dp}{a}( 1/2 + \frac{p}{4b} )  \)

 \(P =  ∫^{min(4b,a)}_0  \frac{dp}{a}( 1/2 + \frac{p}{4b} ) +∫^{a}_{min(4b,a)}  \frac{dp}{a}  \)

\( = [\frac{p}{2a}+ \frac{p^2}{16ba}]^{min(4b,a)}_0 + [\frac{p}{a} ]^{a}_{min(4b,a)} \)

note that if p>4q the probability for a given p is 1

if a<=4b then

\( P= [\frac{p}{2a}+ \frac{p^2}{16ba}]^{a}_0 =\frac{a}{2a}+ \frac{a^2}{16ab} - 0= \frac{1}{2}+ \frac{a}{16b}  \)

if a>4b then

\( P= [\frac{p}{2a}+ \frac{p^2}{16ba}]^{4b}_0 +\frac{a-4b}{a}     =\)

\(    -0 +\frac{4b}{2a}+ \frac{16b^2}{16ab}-0   +\frac{a-4b}{a}=1 -\frac{2b}{2a}=  1- \frac{b}{a}  \)