Solving Quadratic Equations worksheets - Free Printable

To solve the given quadratic equations, we will use either factoring or the quadratic formula. Let's go through each equation step by step.

Quadratic Formula Recap:

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( ax^2 + bx + c = 0 \).

Problem 1: \( x^2 + 13x + 12 = 0 \)

#### Factoring:
We need to find two numbers that multiply to \( 12 \) and add to \( 13 \). These numbers are \( 12 \) and \( 1 \).
\[
x^2 + 13x + 12 = (x + 12)(x + 1) = 0
\]
Setting each factor to zero:
\[
x + 12 = 0 \quad \text{or} \quad x + 1 = 0
\]
\[
x = -12 \quad \text{or} \quad x = -1
\]

#### Solution:
\[
\boxed{x = -12, x = -1}
\]

---

Problem 2: \( 3x^2 - 7x + 4 = 0 \)

#### Factoring:
We need to factor \( 3x^2 - 7x + 4 \). We look for two numbers that multiply to \( 3 \cdot 4 = 12 \) and add to \( -7 \). These numbers are \( -3 \) and \( -4 \).
\[
3x^2 - 7x + 4 = (3x - 4)(x - 1) = 0
\]
Setting each factor to zero:
\[
3x - 4 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = \frac{4}{3} \quad \text{or} \quad x = 1
\]

#### Solution:
\[
\boxed{x = \frac{4}{3}, x = 1}
\]

---

Problem 3: \( x^2 + 5x + 4 = 0 \)

#### Factoring:
We need to find two numbers that multiply to \( 4 \) and add to \( 5 \). These numbers are \( 4 \) and \( 1 \).
\[
x^2 + 5x + 4 = (x + 4)(x + 1) = 0
\]
Setting each factor to zero:
\[
x + 4 = 0 \quad \text{or} \quad x + 1 = 0
\]
\[
x = -4 \quad \text{or} \quad x = -1
\]

#### Solution:
\[
\boxed{x = -4, x = -1}
\]

---

Problem 4: \( 2x^2 + 4x + 3 = 0 \)

#### Using the Quadratic Formula:
Here, \( a = 2 \), \( b = 4 \), and \( c = 3 \). The discriminant is:
\[
b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot 3 = 16 - 24 = -8
\]
Since the discriminant is negative, there are no real solutions.

#### Solution:
\[
\boxed{\text{No real solutions}}
\]

---

Problem 5: \( 3x^2 + 8x - 3 = 0 \)

#### Using the Quadratic Formula:
Here, \( a = 3 \), \( b = 8 \), and \( c = -3 \). The discriminant is:
\[
b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100
\]
Using the quadratic formula:
\[
x = \frac{-8 \pm \sqrt{100}}{2 \cdot 3} = \frac{-8 \pm 10}{6}
\]
This gives:
\[
x = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-8 - 10}{6} = \frac{-18}{6} = -3
\]

#### Solution:
\[
\boxed{x = \frac{1}{3}, x = -3}
\]

---

Problem 6: \( x^2 + 10x - 24 = 0 \)

#### Factoring:
We need to find two numbers that multiply to \( -24 \) and add to \( 10 \). These numbers are \( 12 \) and \( -2 \).
\[
x^2 + 10x - 24 = (x + 12)(x - 2) = 0
\]
Setting each factor to zero:
\[
x + 12 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -12 \quad \text{or} \quad x = 2
\]

#### Solution:
\[
\boxed{x = -12, x = 2}
\]

---

Problem 7: \( x^2 + 3x + 3 = 0 \)

#### Using the Quadratic Formula:
Here, \( a = 1 \), \( b = 3 \), and \( c = 3 \). The discriminant is:
\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 3 = 9 - 12 = -3
\]
Since the discriminant is negative, there are no real solutions.

#### Solution:
\[
\boxed{\text{No real solutions}}
\]

---

Problem 8: \( 4x^2 + 6x - 4 = 0 \)

#### Simplifying:
First, divide the entire equation by 2:
\[
2x^2 + 3x - 2 = 0
\]
Now, factor:
We need two numbers that multiply to \( 2 \cdot (-2) = -4 \) and add to \( 3 \). These numbers are \( 4 \) and \( -1 \).
\[
2x^2 + 3x - 2 = (2x - 1)(x + 2) = 0
\]
Setting each factor to zero:
\[
2x - 1 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = \frac{1}{2} \quad \text{or} \quad x = -2
\]

#### Solution:
\[
\boxed{x = \frac{1}{2}, x = -2}
\]

---

Problem 9: \( 4x^2 + 8x = 0 \)

#### Factoring:
Factor out the greatest common factor:
\[
4x(x + 2) = 0
\]
Setting each factor to zero:
\[
4x = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 0 \quad \text{or} \quad x = -2
\]

#### Solution:
\[
\boxed{x = 0, x = -2}
\]

---

Problem 10: \( 4x^2 - 24x + 36 = 0 \)

#### Simplifying:
First, divide the entire equation by 4:
\[
x^2 - 6x + 9 = 0
\]
This is a perfect square trinomial:
\[
(x - 3)^2 = 0
\]
Setting the factor to zero:
\[
x - 3 = 0
\]
\[
x = 3
\]

#### Solution:
\[
\boxed{x = 3}
\]

---

Final Answers:

1. \( \boxed{x = -12, x = -1} \)
2. \( \boxed{x = \frac{4}{3}, x = 1} \)
3. \( \boxed{x = -4, x = -1} \)
4. \( \boxed{\text{No real solutions}} \)
5. \( \boxed{x = \frac{1}{3}, x = -3} \)
6. \( \boxed{x = -12, x = 2} \)
7. \( \boxed{\text{No real solutions}} \)
8. \( \boxed{x = \frac{1}{2}, x = -2} \)
9. \( \boxed{x = 0, x = -2} \)
10. \( \boxed{x = 3} \)