If `m_1 and m_2` are roots of equation `x^2+(sqrt(3)+2)x+sqrt(3)-1=0` the the area of the triangle formed by lines ` y=m_1x, y = m_2x, y=c` is:

Answer

To find the area of the triangle formed by the lines \(y = m_1 x\), \(y = m_2 x\), and \(y = c\), where \(m_1\) and \(m_2\) are the roots of the equation \(x^2 + (\sqrt{3} + 2)x + (\sqrt{3} - 1) = 0\), we can follow these steps:

### Step 1: Find the roots \(m_1\) and \(m_2\)
We will use the quadratic formula to find the roots of the equation:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
where \(D = b^2 - 4ac\).

Here, \(a = 1\), \(b = \sqrt{3} + 2\), and \(c = \sqrt{3} - 1\).

First, calculate the discriminant \(D\):
\[ D = (\sqrt{3} + 2)^2 - 4 \cdot 1 \cdot (\sqrt{3} - 1) \]
Calculating \(D\):
\[ D = (3 + 4\sqrt{3} + 4) - (4\sqrt{3} - 4) = 7 + 4 \]
\[ D = 11 \]

Now, substituting back into the quadratic formula:
\[ m_1, m_2 = \frac{-(\sqrt{3} + 2) \pm \sqrt{11}}{2} \]

### Step 2: Coordinates of the points of intersection
The points of intersection of the lines with \(y = c\) can be found by setting \(y = c\) in the equations of the lines:
- For \(y = m_1 x\):
\[ c = m_1 x \implies x = \frac{c}{m_1} \implies \text{Point Q} = \left(\frac{c}{m_1}, c\right) \]
- For \(y = m_2 x\):
\[ c = m_2 x \implies x = \frac{c}{m_2} \implies \text{Point R} = \left(\frac{c}{m_2}, c\right) \]
- The third point \(P\) is the origin: \(P = (0, 0)\).

### Step 3: Area of the triangle
The area \(A\) of the triangle formed by points \(P\), \(Q\), and \(R\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| 0(c - c) + \frac{c}{m_1}(c - 0) + \frac{c}{m_2}(0 - c) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| \frac{c^2}{m_1} - \frac{c^2}{m_2} \right| = \frac{c^2}{2} \left| \frac{1}{m_1} - \frac{1}{m_2} \right| \]
Using the identity \(\frac{1}{m_1} - \frac{1}{m_2} = \frac{m_2 - m_1}{m_1 m_2}\):
\[ A = \frac{c^2}{2} \cdot \frac{m_2 - m_1}{m_1 m_2} \]

### Step 4: Calculate \(m_1 - m_2\) and \(m_1 m_2\)
From the quadratic equation:
- The sum of the roots \(m_1 + m_2 = -(\sqrt{3} + 2)\)
- The product of the roots \(m_1 m_2 = \sqrt{3} - 1\)

### Step 5: Substitute values into the area formula
Substituting \(m_1 - m_2\) and \(m_1 m_2\) into the area formula:
\[ A = \frac{c^2}{2} \cdot \frac{\sqrt{11}}{\sqrt{3} - 1} \]
Rationalizing the denominator:
\[ A = \frac{c^2}{2} \cdot \frac{\sqrt{11}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{c^2 \sqrt{11}(\sqrt{3} + 1)}{2(3 - 1)} = \frac{c^2 \sqrt{11}(\sqrt{3} + 1)}{4} \]

### Final Answer
The area of the triangle is:
\[ A = \frac{c^2 \sqrt{11}(\sqrt{3} + 1)}{4} \]