If an equation is real and distinct, you need to show that

the discriminant b2-4ac  is greater than 0 (positive) for any value of x.

$\displaystyle 2{{x}^{2}}+px+p-3=0$

a=2  b=p  c=p-3, substitute into b2-4ac.

$\displaystyle {{(p)}^{2}}-4(2)(p-3)$

$\displaystyle {{p}^{2}}-8(p-3)$

$\displaystyle {{p}^{2}}-8p+24$

Use Completing the Squares

$\displaystyle \pm {{(\frac{b}{2})}^{2}}\Rightarrow \pm {{(\frac{{-8}}{2})}^{2}}\Rightarrow \pm {{(-4)}^{2}}$

$\displaystyle \ {{p}^{2}}-8p+{{(-4)}^{2}}-{{(-4)}^{2}}+24$

$\displaystyle {{(p-4)}^{2}}-{{(-4)}^{2}}+24$

$\displaystyle {{(p-4)}^{2}}+8$

$\displaystyle \text{Since}\ {{(p-4)}^{2}}\ge 0$

$\displaystyle \text{Therefore}\ {{(p-4)}^{2}}+8\ge 0\ \ \text{(Shown)}$

A common mistake is to declare that b2-4ac >0 at the beginning. You should not do that because the question requires you to SHOW i.e. proof that the equation is >0. You should only  declare at the end of the working. Some of my students DID NOT USE completing the squares, as such they could not SHOW that the equation is positive (greater than 0) for all values of x.

additional math (amath) singapore syllabus in johor bahru for sec 3, sec 4 and sec 5 students.