2х²+3х-5=0

To solve the quadratic equation \(2x^2 + 3x - 5 = 0\), we can use the quadratic formula:

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

Here, \(a = 2\), \(b = 3\), and \(c = -5\).

1. Calculate the discriminant \(b^2 - 4ac\):

\[ b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49 \]

2. Now apply the quadratic formula:

\[ x = \frac{{-3 \pm \sqrt{49}}}{{2 \cdot 2}} = \frac{{-3 \pm 7}}{4} \]

3. Calculate the two possible solutions:

\[ x_1 = \frac{{-3 + 7}}{4} = \frac{4}{4} = 1 \]

\[ x_2 = \frac{{-3 - 7}}{4} = \frac{{-10}}{4} = -\frac{5}{2} \]

Thus, the solutions to the equation \(2x^2 + 3x - 5 = 0\) are:

\[ x = 1 \quad \text{and} \quad x = -\frac{5}{2} \]