When does the quadratic equation $a{{x}^{2}}+bx+c=0$ has two distinct real roots?
(a) a = 0
(b) ${{b}^{2}}-4ac$= 0
(c) ${{b}^{2}}-4ac$ < 0
(d) ${{b}^{2}}-4ac$> 0

Hint: You should know that for calculating the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ formula is $\dfrac{-b\pm \sqrt{D}}{2a}$ , where D is the discriminant of the quadratic equation, You should also keep in mind that if the value of D is positive or 0 then roots will be real otherwise roots will be imaginary.

Complete step-by-step answer:
We know that,
For quadratic equation $a{{x}^{2}}+bx+c=0$ roots can be found using the formula,
$\dfrac{-b\pm \sqrt{D}}{2a}$,
Where D is the discriminant of the quadratic equation,
We need to first learn what discriminant is in order to learn about roots of a quadratic equation, so given below is the definition of discriminant,
A discriminant is a value calculated from a quadratic equation. It is used to 'discriminate' between the roots (or solutions) of a quadratic equation.
And the value of the discriminant is given by,
$D={{b}^{2}}-4ac$
There are total of three cases that we need to discuss,
Case 1:
${{b}^{2}}-4ac$ > 0
when the discriminant is greater than zero, this means that the quadratic equation has two real roots, and we will get distinct (different) roots.

Case 2:
${{b}^{2}}-4ac$ > 0
If the discriminant is less than zero, this means that the quadratic equation has no real roots, it means that we will get imaginary roots.
Case 3:
${{b}^{2}}-4ac$ = 0
And, if the discriminant is equal to zero, this means that the quadratic equation has two real roots, but we will get identical roots.
Now, we will come back to the question, it is asked that quadratic equation $a{{x}^{2}}+bx+c=0$ should have two distinct and real roots,
Above mentioned criteria is satisfied by case 1 which is ${{b}^{2}}-4ac$ > 0 , hence
Out of given options, option (d) is correct.

Note: You should try to analyse what is happening here and why the roots are imaginary in the case 2 and why real in case 1, as in case 2 if D is negative value of $\sqrt{D}$ will be negative which in turn will give imaginary root and if D is positive value of $\sqrt{D}$ and hence, roots will be real.
Above analysis will help you remember all of the cases.