3. An equation is shown. \[ 2 x^{2}-5 x-3=0 \] What values of x make the equation true? \( x=\square \) \( x=\square \)

To solve the quadratic equation \( 2x^{2}-5x-3=0 \), we can use the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -5 \), and \( c = -3 \). Plugging these values into the formula gives us: 1. Calculate the discriminant: \( b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49 \). 2. Apply the quadratic formula: \( x = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4} \). This results in two solutions: - \( x = \frac{12}{4} = 3 \) - \( x = \frac{-2}{4} = -\frac{1}{2} \) So, the values of \( x \) that make the equation true are \( x=3 \) and \( x=-\frac{1}{2} \). Don't forget to check your answers! Inserting \( x = 3 \) and \( x = -\frac{1}{2} \) back into the original equation should yield \( 0 \). If they do, you’ve got it right! It’s always a good practice to verify your solutions, just for that extra peace of mind.