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 \( 5x^{2}-13x+6 = 0 \), you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 5 \), \( b = -13 \), and \( c = 6 \). Plugging in these values, you have: \[ x = \frac{-(-13) \pm \sqrt{(-13)^{2} - 4 \cdot 5 \cdot 6}}{2 \cdot 5} \] \[ x = \frac{13 \pm \sqrt{169 - 120}}{10} \] \[ x = \frac{13 \pm \sqrt{49}}{10} \] \[ x = \frac{13 \pm 7}{10} \] Calculating the two possible values: 1. \( x = \frac{20}{10} = 2 \) 2. \( x = \frac{6}{10} = 0.6 \) So, the possible values of \( x \) are \( 2 \) and \( 0.6 \)! --- Understanding quadratics is crucial, as they show up in various fields like physics, biology, and finance! For example, they can model projectile motion or optimize profits. It's like a secret tool used by businesses and scientists alike, making them more relatable to our everyday experiences! When solving quadratics, remember to check for mistakes like sign errors or miscalculating the discriminant. Always confirm that your factors multiply back to the original equation. More often than not, it's small slips that trip you up, so double-checking your work is key to mastering problem-solving in math!