We are asked to determine the type of solutions the quadratic equation $9x^2 + 14x + 5 = 0$ has by using the discriminant. We are not required to solve the equation.

1. Problem Description

We are asked to determine the type of solutions the quadratic equation $9x^2 + 14x + 5 = 0$ has by using the discriminant. We are not required to solve the equation.

2. Solution Steps

The general form of a quadratic equation is $ax^2 + bx + c = 0$.

The discriminant, denoted by $\Delta$, is given by the formula:

$\Delta = b^2 - 4ac$

In the given equation $9x^2 + 14x + 5 = 0$, we have $a = 9$, $b = 14$, and $c = 5$.

Now, we compute the discriminant:

$\Delta = (14)^2 - 4(9)(5)$

$\Delta = 196 - 180$

$\Delta = 16$

Since $\Delta > 0$, the equation has two distinct real solutions.

Since $\Delta = 16$ is a perfect square, the solutions are rational.

If $\Delta > 0$ and $\Delta$ is a perfect square, then the equation has two distinct rational solutions.

If $\Delta > 0$ and $\Delta$ is not a perfect square, then the equation has two distinct irrational solutions.

If $\Delta = 0$, then the equation has one rational solution (a repeated root).

If $\Delta < 0$, then the equation has two nonreal complex solutions.

In our case, $\Delta = 16 > 0$ and is a perfect square ($4^2 = 16$). Thus, the given quadratic equation has two rational solutions.

3. Final Answer

Two rational solutions