The discriminant of a quadratic equation can be calculated using the formula,

Discriminant = b^{2} – 4ac.

The discriminant of a quadratic equation gives us information about the roots of a quadratic equation without explicitly calculating them.

It tells us whether the quadratic equation has two distinct roots, one single root, or no roots at all depending on the sign of the discriminant.

**Discriminant Formula:**

We must first express the quadratic equation in the standard form given as,

ax^{2}+bx+c=0.

Then the discriminant of the quadratic equation is given by the formula,

Discriminant D = b^{2} – 4ac.

Now, depending on the sign of the discriminant we have three possible cases.

**Case 1: Discriminant is Positive**

If the discriminant is strictly positive then it means that the given quadratic equation has two distinct real roots.

Let us understand this by looking at an example. Consider the following quadratic equation.

x^{2}-5x+6=0.

This equation is in the standard form given above and we have that a=1, b= -5, and c=6. Then the discriminant is calculated using the formula,

D = b^{2} – 4ac = (-5)^{2} – 4(1)(6) = 1 > 0.

Since the discriminant is strictly positive we can conclude that the equation has two unequal real roots.

We can verify this by simply solving the equation by the factorization method.

x^{2}-5x+6=0.

x^{2}-3x-2x+6=0.

x(x-3)-2(x-3)=0.

(x-2)(x-3)=0.

x=2, 3.

We see that the equation indeed has two distinct roots 2 and 3.

**Case 2: Discriminant is Zero**

If the discriminant is zero then we conclude that the given quadratic equation has a single real root.

Since a quadratic equation is of degree 2 and is supposed to have two roots we sometimes say that “if the discriminant is zero the quadratic equation has two *repeated* roots”.

**Example:**

Consider the following quadratic equation x^{2}-6x+9=0. Here we have that, a=1, b= -6, and c=9.

D = b^{2} – 4ac = (-6)^{2} – 4(1)(9) = 0.

Let us also verify by factorizing the polynomial that the equation indeed has two repeated roots.

x^{2}-6x+9=0.

x^{2}-3x-3x+9=0.

x(x-3)-3(x-3)=0.

(x-3)(x-3)=0.

x=3, 3. We see that root 3 occurs with a multiplicity of 2.

**Case 3: Discriminant is Negative**

If the discriminant is negative then the quadratic equation has no real roots. This means that the equation cannot be solved.

For example, the discriminant of the quadratic equation, x^{2}+x+1=0 is given as, D = b^{2} – 4ac = (1)^{2} – 4(1)(1) = -3 < 0 .

**Why can a quadratic equation not be solved if the discriminant is negative?**

A quadratic equation does not have roots if the discriminant is negative.

This can be understood by looking at the quadratic formula. The two roots of a quadratic equation are given by the formula, x = (-b ± √D)/2a.

Since it is impossible to take the square root of a real number, if the quantity under the root is negative the equation is impossible to solve.

In order to solve such equations one needs to expand the number system to include *complex numbers*.