An easy one for you

$let\quad the\quad roots\quad of\quad the\quad equation\quad { x }^{ 2 }-px+q=0\quad be\quad \alpha \quad and\quad \beta \\ given\quad that\quad \alpha -\beta =1\quad (considering\quad \alpha >\beta )\\ We\quad can\quad express\quad \alpha -\beta \quad in\quad the\quad form\quad \sqrt { { \left( \alpha +\beta \right) }^{ 2 }-4\alpha \beta } \\ \Rightarrow \quad \alpha -\beta =\sqrt { { \left( \alpha +\beta \right) }^{ 2 }-4\alpha \beta } \\ \Rightarrow 1={ \left( \alpha +\beta \right) }^{ 2 }-4\alpha \beta \\ In\quad a\quad quadratic\quad equation\quad a{ x }^{ 2 }+bx+c=0\\ \alpha +\beta =\dfrac { -b }{ a } \quad and\quad \alpha \beta =\dfrac { c }{ a } \quad \\ \Rightarrow 1={ \left( \dfrac { -\left( -p \right) }{ 1 } \right) }^{ 2 }-4 \left(\dfrac { q }{ 1 }\right) \\ \Rightarrow 1={ p }^{ 2 }-4q\\ \Rightarrow { p }^{ 2 }=4q+1$

$x^2-3x+2=0$ satisfies.

$3^2=2×4+1$