## Answer :

**quadratic equation**.

Let's illustrate the solution with an example:

Consider the quadratic equation [tex]\( 2x^2 + 3x + 1 = 0 \)[/tex] where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 1 \)[/tex].

To find the solutions (roots) of the quadratic equation, we can use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

First, compute the discriminant [tex]\( \Delta \)[/tex] which is given by [tex]\( b^2 - 4ac \)[/tex]:

[tex]\[ \Delta = 3^2 - 4(2)(1) = 9 - 8 = 1 \][/tex]

The discriminant is 1. Since the discriminant is greater than 0, we have two distinct real roots. We can now find the roots.

For the first root:

[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-3 + \sqrt{1}}{2(2)} = \frac{-3 + 1}{4} = \frac{-2}{4} = -0.5 \][/tex]

For the second root:

[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-3 - \sqrt{1}}{2(2)} = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \][/tex]

Thus, the roots of the quadratic equation [tex]\( 2x^2 + 3x + 1 = 0 \)[/tex] are [tex]\( -0.5 \)[/tex] and [tex]\( -1 \)[/tex], and the discriminant is [tex]\( 1 \)[/tex].