Solve the following equation using the quadratic formula: \[ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \] 1st attempt See Periodic Table Express \( x \) values to a minimum of two significant figures. \[ 0=5 x^{2}-2 x-22 \] The \( x \) obtained from the "plus" operation = \( \square \) The \( x \) obtained from the "minus" operation= \( \square \)

To solve the equation \(5x^2 - 2x - 22 = 0\) using the quadratic formula, we first identify the coefficients: \(a = 5\), \(b = -2\), and \(c = -22\). Plugging these values into the formula gives us: \[ x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4 \cdot 5 \cdot (-22)}}{2 \cdot 5} \] Calculating inside the square root: \[ (-2)^{2} = 4 \quad \text{and} \quad -4 \cdot 5 \cdot -22 = 440 \] So it becomes: \[ x = \frac{2 \pm \sqrt{4 + 440}}{10} \] \[ = \frac{2 \pm \sqrt{444}}{10} \] \[ = \frac{2 \pm 21.07}{10} \] Now calculating the two potential \(x\) values: 1. For the "plus" operation: \[ x = \frac{2 + 21.07}{10} = \frac{23.07}{10} = 2.31 \quad (\text{to two significant figures: } 2.3) \] 2. For the "minus" operation: \[ x = \frac{2 - 21.07}{10} = \frac{-19.07}{10} = -1.907 \quad (\text{to two significant figures: } -1.9) \] The values of \(x\) are: The \( x \) obtained from the "plus" operation = \(2.3\) The \( x \) obtained from the "minus" operation= \(-1.9\) This equation not only flexes your mathematical muscles but connects to real-world situations, especially in physics for projectile motion or calculating areas. Plus, practicing with quadratics boosts your problem-solving prowess, keeping your brain sharp!