[tex]ax^2+bx+c=0\\\\\text{From quadratic formula the roots are equal:}\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\ \text{and}\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ 2x^2+7x-15=0\\\\a=2,\ b=7,\ c=-15\\\\\text{and}\ r,\ s\ \text{are two solution}\\\\\sqrt{b^2-4ac}=\sqrt{7^2-4(2)(-15)}=\sqrt{49+120}=\sqrt{169}=13\\\\x_1=\dfrac{-7-13}{2(2)}=\dfrac{-20}{4}=-5\\\\x_2=\dfrac{-7+13}{2(2)}=\dfrac{6}{4}=\dfrac{3}{2}\\\\\dfrac{3}{2}>-5\ \text{therefore}\ r=\dfrac{3}{2}\ \text{and}\ s=-5.[/tex]
[tex]r-s=\dfrac{3}{2}-(-5)}=\dfrac{3}{2}+5=\dfrac{3}{2}+\dfrac{10}{2}=\dfrac{13}{2}\\\\Answer:\ \boxed{B)\ \dfrac{13}{2}}[/tex]
