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 \( x = -0.6 \).

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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.

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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 \( x = 2 \).

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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 \( x = -11 \).

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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 \( x = -2 \).

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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 \( x = -3 \).

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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 \( c = -\frac{1}{7} \).

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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. \( c = -\frac{1}{7} \)