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