The equation $4x^2 - 15x + c = 0$ has a solution of $-\frac{1}{4}$. We need to find the value of $c$ and the other solution to the equation.

1. Problem Description

The equation $4x^2 - 15x + c = 0$ has a solution of $-\frac{1}{4}$. We need to find the value of $c$ and the other solution to the equation.

2. Solution Steps

First, we substitute $x = -\frac{1}{4}$ into the equation $4x^2 - 15x + c = 0$ to solve for $c$.

$4(-\frac{1}{4})^2 - 15(-\frac{1}{4}) + c = 0$

$4(\frac{1}{16}) + \frac{15}{4} + c = 0$

$\frac{1}{4} + \frac{15}{4} + c = 0$

$\frac{16}{4} + c = 0$

$4 + c = 0$

$c = -4$

Now we have the quadratic equation $4x^2 - 15x - 4 = 0$. Since we know one solution is $x = -\frac{1}{4}$, we can find the other solution using the fact that the product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $\frac{c}{a}$. In this case, the product of the roots is $\frac{-4}{4} = -1$.

Let $x_1$ and $x_2$ be the two solutions. Then $x_1 \cdot x_2 = -1$. We are given $x_1 = -\frac{1}{4}$, so

$(-\frac{1}{4}) x_2 = -1$

$x_2 = 4$

Alternatively, we can factor the quadratic equation.

$4x^2 - 15x - 4 = 0$

$4x^2 - 16x + x - 4 = 0$

$4x(x - 4) + 1(x - 4) = 0$

$(4x + 1)(x - 4) = 0$

Thus, $4x + 1 = 0$ or $x - 4 = 0$.

$4x = -1$ so $x = -\frac{1}{4}$

$x = 4$

The two solutions are $x = -\frac{1}{4}$ and $x = 4$.

3. Final Answer

$c = -4$

The other solution is $4$.