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Let's calculate the discriminants for each of the given quadratic equations. The discriminant \(\Delta\) for a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\[
\Delta = b^2 - 4ac
\]

The number of real solutions is determined by the discriminant as follows:
- If \(\Delta > 0\), the quadratic has two distinct real solutions.
- If \(\Delta = 0\), the quadratic has exactly one real solution (a repeated root).
- If \(\Delta < 0\), the quadratic has no real solutions (the solutions are complex).

Now let's analyze each quadratic:

### (a) \(x^2 + 4x - 5 = 0\)
Here, \(a = 1\), \(b = 4\), and \(c = -5\).

\[
\Delta = 4^2 - 4(1)(-5) = 16 + 20 = 36
\]

\(\Delta = 36 > 0\), so this quadratic has **two distinct real solutions**.

### (b) \(2x^2 + x + 6 = 0\)
Here, \(a = 2\), \(b = 1\), and \(c = 6\).

\[
\Delta = 1^2 - 4(2)(6) = 1 - 48 = -47
\]

\(\Delta = -47 < 0\), so this quadratic has **no real solutions** (only complex solutions).

### (c) \(3x^2 - 4x + 2x - 1 = 0\)
First, simplify this to \(3x^2 - 2x - 1 = 0\), where \(a = 3\), \(b = -2\), and \(c = -1\).

\[
\Delta = (-2)^2 - 4(3)(-1) = 4 + 12 = 16
\]

\(\Delta = 16 > 0\), so this quadratic has **two distinct real solutions**.

### (d) \(12 - x - x^2 = 0\)
Rearrange this as \(-x^2 - x + 12 = 0\) or \(x^2 + x - 12 = 0\) by multiplying through by \(-1\). Now, \(a = 1\), \(b = 1\), and \(c = -12\).

\[
\Delta = 1^2 - 4(1)(-12) = 1 + 48 = 49
\]

\(\Delta = 49 > 0\), so this quadratic has **two distinct real solutions**.

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Would you like more details on any part of this? Here are some related questions:

1. How can you solve quadratic equations using the quadratic formula?
2. What is the significance of the discriminant in understanding the nature of solutions?
3. How can you factor quadratic equations when the discriminant is positive?
4. What methods are available to solve quadratic equations without using the discriminant?
5. Can quadratic equations have complex solutions if their discriminants are negative?

**Tip:** Always check the discriminant first to quickly determine the nature of the solutions before proceeding with the complete solving process!

Calculate the discriminant and state the number of real solutions for the following quadratics: (a) x^2 + 4x - 5 = 0, (b) 2x^2 + x + 6 = 0, (c) 3x^2 - 4x + 2x - 1 = 0, (d) 12 - x - x^2 = 0

Learn how to calculate the discriminant for the quadratic equations: x^2 + 4x - 5, 2x^2 + x + 6, and others. This problem covers algebraic concepts like quadratic equations and the discriminant formula. Ideal for high school students.

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