2x^2-3x-7=0 (If there is more than one solution, separate them with commas.) x= square

To solve the quadratic equation \(2x^2 - 3x - 7 = 0\), we can use the quadratic formula:<br /><br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />Here, \(a = 2\), \(b = -3\), and \(c = -7\).<br /><br />First, calculate the discriminant (\(\Delta\)):<br /><br />\[ \Delta = b^2 - 4ac \]<br />\[ \Delta = (-3)^2 - 4(2)(-7) \]<br />\[ \Delta = 9 + 56 \]<br />\[ \Delta = 65 \]<br /><br />Now, substitute \(\Delta\) back into the quadratic formula:<br /><br />\[ x = \frac{-(-3) \pm \sqrt{65}}{2(2)} \]<br />\[ x = \frac{3 \pm \sqrt{65}}{4} \]<br /><br />So, the solutions are:<br /><br />\[ x = \frac{3 + \sqrt{65}}{4}, \quad x = \frac{3 - \sqrt{65}}{4} \]<br /><br />Therefore, the solutions to the equation \(2x^2 - 3x - 7 = 0\) are:<br /><br />\[ x = \frac{3 + \sqrt{65}}{4}, \quad x = \frac{3 - \sqrt{65}}{4} \]