For what values of `x in R`, the following expressions are negative
i) `-6x^(2)+2x -3` ii) `15+4x-3x^(2)` iii) `2x^(2)+5x -3` iv) `x^(2)-7x+10`

Answer

To determine the values of \( x \in \mathbb{R} \) for which the given expressions are negative, we will analyze each expression one by one, finding the roots and determining the intervals where the expressions are negative.

### i) Expression: \(-6x^2 + 2x - 3\)

1. **Find the discriminant (D)**:
\[ D = b^2 - 4ac = 2^2 - 4(-6)(-3) = 4 - 72 = -68 \]
Since \( D < 0 \), there are no real roots.

2. **Conclusion**: The expression \(-6x^2 + 2x - 3\) is always negative for all \( x \in \mathbb{R} \).

### ii) Expression: \(15 + 4x - 3x^2\)

1. **Rearranging**:
\[ -3x^2 + 4x + 15 \]

2. **Find the discriminant (D)**:
\[ D = b^2 - 4ac = 4^2 - 4(-3)(15) = 16 + 180 = 196 \]
Since \( D > 0 \), there are two real roots.

3. **Finding the roots**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{196}}{2(-3)} = \frac{-4 \pm 14}{-6} \]
Roots:
\[ x_1 = \frac{10}{-6} = -\frac{5}{3}, \quad x_2 = \frac{-18}{-6} = 3 \]

4. **Determine intervals**: The quadratic opens downwards (since the coefficient of \( x^2 \) is negative). The expression is negative between the roots:
\[ -\frac{5}{3} < x < 3 \]

### iii) Expression: \(2x^2 + 5x - 3\)

1. **Find the discriminant (D)**:
\[ D = b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49 \]
Since \( D > 0 \), there are two real roots.

2. **Finding the roots**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm 7}{4} \]
Roots:
\[ x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-12}{4} = -3 \]

3. **Determine intervals**: The quadratic opens upwards (since the coefficient of \( x^2 \) is positive). The expression is negative between the roots:
\[ -3 < x < \frac{1}{2} \]

### iv) Expression: \(x^2 - 7x + 10\)

1. **Find the discriminant (D)**:
\[ D = b^2 - 4ac = (-7)^2 - 4(1)(10) = 49 - 40 = 9 \]
Since \( D > 0 \), there are two real roots.

2. **Finding the roots**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm 3}{2} \]
Roots:
\[ x_1 = \frac{10}{2} = 5, \quad x_2 = \frac{4}{2} = 2 \]

3. **Determine intervals**: The quadratic opens upwards (since the coefficient of \( x^2 \) is positive). The expression is negative between the roots:
\[ 2 < x < 5 \]

### Summary of Results
1. For \(-6x^2 + 2x - 3\): Always negative.
2. For \(15 + 4x - 3x^2\): \(-\frac{5}{3} < x < 3\).
3. For \(2x^2 + 5x - 3\): \(-3 < x < \frac{1}{2}\).
4. For \(x^2 - 7x + 10\): \(2 < x < 5\).