2х+3х^2-1=5

To solve the equation \(2x + 3x^2 - 1 = 5\), we first rearrange it so that all terms are on one side:

\[ 3x^2 + 2x - 1 - 5 = 0 \]

This simplifies to:

\[ 3x^2 + 2x - 6 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula, which is given by:

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

In this case, \(a = 3\), \(b = 2\), and \(c = -6\).

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

\[ b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot (-6) = 4 + 72 = 76 \]

Since the discriminant is positive, there are two real solutions. Now we substitute the values into the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{76}}{2 \cdot 3} = \frac{-2 \pm 2\sqrt{19}}{6} = \frac{-1 \pm \sqrt{19}}{3} \]

Thus, the solutions to the equation \(2x + 3x^2 - 1 = 5\) are:

\[ x = \frac{-1 + \sqrt{19}}{3} \quad \text{and} \quad x = \frac{-1 - \sqrt{19}}{3} \]

These represent the two values of \(x\) that satisfy the original equation.