How many real number or complex number roots does the following equation have? 

y = x² + 4x + 1

There's an easier way to find out than going through the trouble of finding all the roots yourself.

Let's use the discriminant: b² – 4ac.

Er…we might need to know what a, b, and c are first.

a = 1
b = 4
c = 1

b² – 4ac = 16 – 4(1)(1) = 16 – 4 = 12

The discriminant is positive, so there are two real number roots for this equation.

How many real number or complex number roots does the following equation have? 

y = x² – x + 3

Let's start off with the setup.

a = 1
b = -1
c = 3

And now we plug away.

b² – 4ac = (-1)² – 4(1)(3) = 1 – 12 = -11

This time the discriminant is negative, so there are two complex number roots (and no real number roots).

How many real number or complex number roots does the following equation have?

y = -2x² – 4x + 1

a = -2
b = -4
c = 1

b² – 4ac = (-4)² – 4(-2)(1) = 16 + 8 = 24

A positive outcome leads to real results—and two real roots.

How many real number or complex number roots does the following equation have?

y = -9x² + 6x – 1

a = -9
b = 6
c = -1

b² – 4ac = (6)² – 4(-9)(-1) = 36 – 36 = 0

The discriminant…it's nothing. A nihilist's dream. In the end, there's only one (real root).