Find the zeros for each of the following quadratics. If you use the quadratic formula please report \( D \) and write out your formulas. Please round decimal answers to two decimal places. 1. \( x^{2}-12 x+32 \) Solve the following equations. 3. \( x^{2}+12 x=-11 \) 3. Thes of quccircitic \( 6 x^{2}-7 x=14 \)

### Problem 1: Find the zeros of \( x^{2} - 12x + 32 \) 1. Write the equation in standard form: \[ x^2 - 12x + 32 = 0 \] 2. We attempt to factor the quadratic. Look for two numbers whose product is \(32\) and whose sum is \(-12\). These numbers are \(-4\) and \(-8\). 3. Factor the quadratic: \[ (x - 4)(x - 8) = 0 \] 4. Set each factor equal to zero and solve for \( x \): \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] \[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \] 5. **Zeros:** \( x = 4 \) and \( x = 8 \). --- ### Problem 2: Solve the equation \( x^{2} + 12x = -11 \) 1. Bring all terms to one side to write the equation in standard form: \[ x^{2} + 12x + 11 = 0 \] 2. Compute the discriminant \((D)\): \[ D = b^2 - 4ac = 12^2 - 4(1)(11) = 144 - 44 = 100 \] 3. Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{100}}{2} = \frac{-12 \pm 10}{2} \] 4. Solve for \( x \): \[ x = \frac{-12 + 10}{2} = \frac{-2}{2} = -1 \] \[ x = \frac{-12 - 10}{2} = \frac{-22}{2} = -11 \] 5. **Solutions:** \( x = -1 \) and \( x = -11 \). --- ### Problem 3: Solve the equation \( 6x^2 - 7x = 14 \) 1. Rearrange the equation to set it equal to zero: \[ 6x^2 - 7x - 14 = 0 \] 2. Compute the discriminant \((D)\): \[ D = b^2 - 4ac = (-7)^2 - 4(6)(-14) = 49 + 336 = 385 \] 3. Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{385}}{12} \] 4. Approximate the square root and solutions: - Calculate \(\sqrt{385}\): \[ \sqrt{385} \approx 19.62 \] - First solution: \[ x \approx \frac{7 + 19.62}{12} = \frac{26.62}{12} \approx 2.22 \] - Second solution: \[ x \approx \frac{7 - 19.62}{12} = \frac{-12.62}{12} \approx -1.05 \] 5. **Solutions:** \( x \approx 2.22 \) and \( x \approx -1.05 \).