5х^2+9х-2<0

To solve the inequality \( 5x^2 + 9x - 2 < 0 \), we first need to find the roots of the corresponding quadratic equation \( 5x^2 + 9x - 2 = 0 \).

We can use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 5 \), \( b = 9 \), and \( c = -2 \).

First, we calculate the discriminant \( b^2 - 4ac \):

\[ b^2 - 4ac = 9^2 - 4(5)(-2) = 81 + 40 = 121 \]

Now we can find the roots:

\[ x = \frac{-9 \pm \sqrt{121}}{2 \cdot 5} = \frac{-9 \pm 11}{10} \]

Calculating the two roots:

1. For \( x = \frac{-9 + 11}{10} = \frac{2}{10} = \frac{1}{5} \)
2. For \( x = \frac{-9 - 11}{10} = \frac{-20}{10} = -2 \)

Thus, the roots of the equation are \( x = -2 \) and \( x = \frac{1}{5} \).

Now we need to analyze the intervals defined by these roots: \( (-\infty, -2) \), \( (-2, \frac{1}{5}) \), and \( (\frac{1}{5}, \infty) \).

We will test a number from each interval in the inequality \( 5x^2 + 9x - 2 < 0 \):

1. For \( x = -3 \) (in the interval \( (-\infty, -2) \)): \[ 5(-3)^2 + 9(-3) - 2 = 45 - 27 - 2 = 16 \quad (\text{not less than } 0) \]

2. For \( x = 0 \) (in the interval \( (-2, \frac{1}{5}) \)): \[ 5(0)^2 + 9(0) - 2 = -2 \quad (\text{less than } 0) \]

3. For \( x = 1 \) (in the interval \( (\frac{1}{5}, \infty) \)): \[ 5(1)^2 + 9(1) - 2 = 5 + 9 - 2 = 12 \quad (\text{not less than } 0) \]

Now, we summarize the results:

• The quadratic expression is positive in the intervals \( (-\infty, -2) \) and \( (\frac{1}{5}, \infty) \).
• The quadratic expression is negative in the interval \( (-2, \frac{1}{5}) \).

Finally, the solution to the inequality \( 5x^2 + 9x - 2 < 0 \) is:

\[ \boxed{(-2, \frac{1}{5})} \]