If \( 5 x^{2}-13 x+6 \) equals zero, what are the possible values of \( x \) ?

To find the possible values of \( x \) in the equation \( 5x^{2} - 13x + 6 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \( a = 5 \), \( b = -13 \), and \( c = 6 \). Plugging in these values gives: \[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 5 \cdot 6}}{2 \cdot 5} \] \[ x = \frac{13 \pm \sqrt{169 - 120}}{10} \] \[ x = \frac{13 \pm \sqrt{49}}{10} \] \[ x = \frac{13 \pm 7}{10} \] This results in two possible solutions: 1. \( x = \frac{20}{10} = 2 \) 2. \( x = \frac{6}{10} = \frac{3}{5} \) Therefore, the possible values of \( x \) are \( x = 2 \) and \( x = \frac{3}{5} \). If you ever find yourself dealing with quadratic equations, remember that they can often have either two distinct solutions, one double solution or no real solution at all! Always check the discriminant \( b^2 - 4ac \) to determine the nature of the roots. Additionally, once you've found the roots, you can quickly verify them by plugging them back into the original equation. It’s a real joy to see your solutions satisfy the equation!