4x^2-3x-5=0

To solve the quadratic equation \(4x^2 - 3x - 5 = 0\), we can use the quadratic formula:

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

Here, the coefficients are:

• \(a = 4\)
• \(b = -3\)
• \(c = -5\)

Now, we can substitute these values into the formula. First, we calculate the discriminant \(b^2 - 4ac\):

\[ b^2 - 4ac = (-3)^2 - 4 \cdot 4 \cdot (-5) = 9 + 80 = 89 \]

Now, substituting the values into the quadratic formula:

\[ x = \frac{-(-3) \pm \sqrt{89}}{2 \cdot 4} = \frac{3 \pm \sqrt{89}}{8} \]

Thus, the solutions to the equation \(4x^2 - 3x - 5 = 0\) are:

\[ x = \frac{3 + \sqrt{89}}{8} \quad \text{and} \quad x = \frac{3 - \sqrt{89}}{8} \]