Score: \( 1 / 5 \) Finding Zeros with Technology Question Find the zeros of the function \( f(x)=2 x^{2}-23.6 x+67 \). Round values to the nearest thousandths (if necessary). Answer Attempt 1 out of 2 \( \oplus \) Add a Zero \( \Theta \) No Zeros \( x=\square \) Submit Answer

Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}-23.6x+67=0\) - step1: Convert the decimal into a fraction: \(2x^{2}-\frac{118}{5}x+67=0\) - step2: Multiply both sides: \(5\left(2x^{2}-\frac{118}{5}x+67\right)=5\times 0\) - step3: Calculate: \(10x^{2}-118x+335=0\) - step4: Solve using the quadratic formula: \(x=\frac{118\pm \sqrt{\left(-118\right)^{2}-4\times 10\times 335}}{2\times 10}\) - step5: Simplify the expression: \(x=\frac{118\pm \sqrt{\left(-118\right)^{2}-4\times 10\times 335}}{20}\) - step6: Simplify the expression: \(x=\frac{118\pm \sqrt{524}}{20}\) - step7: Simplify the expression: \(x=\frac{118\pm 2\sqrt{131}}{20}\) - step8: Separate into possible cases: \(\begin{align}&x=\frac{118+2\sqrt{131}}{20}\\&x=\frac{118-2\sqrt{131}}{20}\end{align}\) - step9: Simplify the expression: \(\begin{align}&x=\frac{59+\sqrt{131}}{10}\\&x=\frac{118-2\sqrt{131}}{20}\end{align}\) - step10: Simplify the expression: \(\begin{align}&x=\frac{59+\sqrt{131}}{10}\\&x=\frac{59-\sqrt{131}}{10}\end{align}\) - step11: Rewrite: \(x_{1}=\frac{59-\sqrt{131}}{10},x_{2}=\frac{59+\sqrt{131}}{10}\) Calculate or simplify the expression \( (59+\sqrt(131))/10 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(59+\sqrt{131}\right)}{10}\) - step1: Remove the parentheses: \(\frac{59+\sqrt{131}}{10}\) Calculate or simplify the expression \( (59-\sqrt(131))/10 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(59-\sqrt{131}\right)}{10}\) - step1: Remove the parentheses: \(\frac{59-\sqrt{131}}{10}\) To find the zeros of the function \( f(x) = 2x^2 - 23.6x + 67 \), we need to solve the equation \( 2x^2 - 23.6x + 67 = 0 \). ### Step 1: Identify the coefficients The coefficients of the quadratic equation are: - \( a = 2 \) - \( b = -23.6 \) - \( c = 67 \) ### Step 2: Use the quadratic formula The quadratic formula to find the roots (zeros) of the equation \( ax^2 + bx + c = 0 \) is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] ### Step 3: Calculate the discriminant First, we calculate the discriminant \( D = b^2 - 4ac \): \[ D = (-23.6)^2 - 4 \cdot 2 \cdot 67 \] Calculating this gives: \[ D = 556.96 - 536 = 20.96 \] ### Step 4: Calculate the zeros Now we can find the zeros using the quadratic formula: \[ x_{1} = \frac{23.6 - \sqrt{20.96}}{4} \] \[ x_{2} = \frac{23.6 + \sqrt{20.96}}{4} \] ### Step 5: Evaluate the zeros Calculating these values gives: - For \( x_{1} \): \[ x_{1} \approx 4.755 \] - For \( x_{2} \): \[ x_{2} \approx 7.045 \] ### Final Answer The zeros of the function \( f(x) = 2x^2 - 23.6x + 67 \) are approximately: - \( x \approx 4.755 \) - \( x \approx 7.045 \) You can submit these values as your answer.