1) 5x ^ 2 = 22x + 15 2) 3x ^ 2 + 9 = 10x 3) x ^ 2 - 2x = 0 4) 121 - x ^ 2 = 0 5) 3x - 6 + 3x ^ 2 = 0 6) 2x ^ 2 = 4x + 30 7) 14c + 49c ^ 2 + 1 = 0

Let's solve each of the equations one by one. ### 1) \( 5x^2 = 22x + 15 \) Rearranging the equation gives: \[ 5x^2 - 22x - 15 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5 \), \( b = -22 \), and \( c = -15 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (-22)^2 - 4(5)(-15) = 484 + 300 = 784 \] 2. Now, apply the quadratic formula: \[ x = \frac{22 \pm \sqrt{784}}{2 \cdot 5} = \frac{22 \pm 28}{10} \] This gives us two solutions: - \( x = \frac{50}{10} = 5 \) - \( x = \frac{-6}{10} = -0.6 \) So, the solutions are \( x = 5 \) and \( x = -0.6 \). --- ### 2) \( 3x^2 + 9 = 10x \) Rearranging gives: \[ 3x^2 - 10x + 9 = 0 \] Using the quadratic formula where \( a = 3 \), \( b = -10 \), and \( c = 9 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (-10)^2 - 4(3)(9) = 100 - 108 = -8 \] Since the discriminant is negative, there are no real solutions. --- ### 3) \( x^2 - 2x = 0 \) Factoring out \( x \): \[ x(x - 2) = 0 \] Setting each factor to zero gives us: - \( x = 0 \) - \( x = 2 \) So, the solutions are \( x = 0 \) and \( x = 2 \). --- ### 4) \( 121 - x^2 = 0 \) Rearranging gives: \[ x^2 = 121 \] Taking the square root of both sides: \[ x = \pm 11 \] So, the solutions are \( x = 11 \) and \( x = -11 \). --- ### 5) \( 3x - 6 + 3x^2 = 0 \) Rearranging gives: \[ 3x^2 + 3x - 6 = 0 \] Dividing the entire equation by 3: \[ x^2 + x - 2 = 0 \] Factoring gives: \[ (x - 1)(x + 2) = 0 \] Setting each factor to zero yields: - \( x = 1 \) - \( x = -2 \) So, the solutions are \( x = 1 \) and \( x = -2 \). --- ### 6) \( 2x^2 = 4x + 30 \) Rearranging gives: \[ 2x^2 - 4x - 30 = 0 \] Dividing through by 2: \[ x^2 - 2x - 15 = 0 \] Factoring gives: \[ (x - 5)(x + 3) = 0 \] Setting each factor to zero gives: - \( x = 5 \) - \( x = -3 \) So, the solutions are \( x = 5 \) and \( x = -3 \). --- ### 7) \( 14c + 49c^2 + 1 = 0 \) Rearranging gives: \[ 49c^2 + 14c + 1 = 0 \] Using the quadratic formula where \( a = 49 \), \( b = 14 \), and \( c = 1 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (14)^2 - 4(49)(1) = 196 - 196 = 0 \] Since the discriminant is zero, we have one real solution: \[ c = \frac{-b}{2a} = \frac{-14}{2 \cdot 49} = \frac{-14}{98} = -\frac{1}{7} \] So, the solution is \( c = -\frac{1}{7} \). --- ### Summary of Solutions: 1. \( x = 5, -0.6 \) 2. No real solutions 3. \( x = 0, 2 \) 4. \( x = 11, -11 \) 5. \( x = 1, -2 \) 6. \( x = 5, -3 \) 7. \( c = -\frac{1}{7} \)