The problem asks us to determine the type of solutions of the quadratic equation $4x(x-1) + 14 = 10$ using the discriminant, without actually solving the equation. The possible solution types are: two rational solutions, two irrational solutions, one rational solution, or two nonreal complex solutions.

1. Problem Description

The problem asks us to determine the type of solutions of the quadratic equation $4x(x-1) + 14 = 10$ using the discriminant, without actually solving the equation. The possible solution types are: two rational solutions, two irrational solutions, one rational solution, or two nonreal complex solutions.

2. Solution Steps

First, we need to rewrite the given equation in the standard quadratic form $ax^2 + bx + c = 0$.

$4x(x-1) + 14 = 10$

$4x^2 - 4x + 14 = 10$

$4x^2 - 4x + 14 - 10 = 0$

$4x^2 - 4x + 4 = 0$

Now, we can identify the coefficients: $a = 4$, $b = -4$, and $c = 4$.

The discriminant is given by the formula:

$D = b^2 - 4ac$

Substitute the values of $a$, $b$, and $c$ into the formula:

$D = (-4)^2 - 4(4)(4)$

$D = 16 - 64$

$D = -48$

Now, let's analyze the discriminant:

- If $D > 0$, there are two distinct real solutions.

- If $D$ is a perfect square, the solutions are rational.

- If $D$ is not a perfect square, the solutions are irrational.

- If $D = 0$, there is one real solution (a repeated root), which is rational.

- If $D < 0$, there are two nonreal complex solutions (conjugate pairs).

In our case, $D = -48$. Since $D < 0$, the equation has two nonreal complex solutions.

3. Final Answer

Two nonreal complex solutions