3x2+ 8x - 3= 0;

To solve the quadratic equation \(3x^2 + 8x - 3 = 0\), you can use the quadratic formula:

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

In this case, the coefficients are:

• \(a = 3\)
• \(b = 8\)
• \(c = -3\)

Now, plug these values into the formula:

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

\[ b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100 \]

2. Now substitute back into the quadratic formula:

\[ x = \frac{-8 \pm \sqrt{100}}{2 \cdot 3} \]

3. Calculate \(\sqrt{100}\):

\[ \sqrt{100} = 10 \]

4. Now substitute this value in:

\[ x = \frac{-8 \pm 10}{6} \]

This gives us two possible solutions:

1. \(x = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}\)

2. \(x = \frac{-8 - 10}{6} = \frac{-18}{6} = -3\)

So the solutions to the equation \(3x^2 + 8x - 3 = 0\) are:

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