Quadratic Formula Worksheets - Math Monks - Free Printable

To solve the given quadratic equations, we will use various methods such as factoring, completing the square, or applying the quadratic formula. Let's solve each equation step by step.

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Equation 1: \( 10x^2 - 26x + 12 = 0 \)

#### Step 1: Simplify if possible
The equation is already in standard form \( ax^2 + bx + c = 0 \), where \( a = 10 \), \( b = -26 \), and \( c = 12 \).

#### Step 2: Factorize
We look for two numbers that multiply to \( 10 \times 12 = 120 \) and add up to \( -26 \). These numbers are \( -20 \) and \( -6 \).

Rewrite the middle term:
\[
10x^2 - 26x + 12 = 10x^2 - 20x - 6x + 12
\]

Factor by grouping:
\[
= 2x(5x - 10) - 3(5x - 10)
\]
\[
= (2x - 3)(5x - 4)
\]

#### Step 3: Solve for \( x \)
Set each factor equal to zero:
\[
2x - 3 = 0 \quad \text{or} \quad 5x - 4 = 0
\]
\[
2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}
\]
\[
5x = 4 \quad \Rightarrow \quad x = \frac{4}{5}
\]

#### Solution:
\[
x = \frac{3}{2}, \, x = \frac{4}{5}
\]

---

Equation 2: \( 3x^2 + 14x - 49 = 0 \)

#### Step 1: Use the quadratic formula
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = 14 \), and \( c = -49 \).

Calculate the discriminant:
\[
b^2 - 4ac = 14^2 - 4 \cdot 3 \cdot (-49) = 196 + 588 = 784
\]

Take the square root of the discriminant:
\[
\sqrt{784} = 28
\]

Apply the quadratic formula:
\[
x = \frac{-14 \pm 28}{2 \cdot 3} = \frac{-14 \pm 28}{6}
\]

Solve for the two roots:
\[
x = \frac{-14 + 28}{6} = \frac{14}{6} = \frac{7}{3}
\]
\[
x = \frac{-14 - 28}{6} = \frac{-42}{6} = -7
\]

#### Solution:
\[
x = \frac{7}{3}, \, x = -7
\]

---

Equation 3: \( 2x^2 + 3x - 20 = 0 \)

#### Step 1: Factorize
We look for two numbers that multiply to \( 2 \times (-20) = -40 \) and add up to \( 3 \). These numbers are \( 8 \) and \( -5 \).

Rewrite the middle term:
\[
2x^2 + 3x - 20 = 2x^2 + 8x - 5x - 20
\]

Factor by grouping:
\[
= 2x(x + 4) - 5(x + 4)
\]
\[
= (2x - 5)(x + 4)
\]

#### Step 2: Solve for \( x \)
Set each factor equal to zero:
\[
2x - 5 = 0 \quad \text{or} \quad x + 4 = 0
\]
\[
2x = 5 \quad \Rightarrow \quad x = \frac{5}{2}
\]
\[
x = -4
\]

#### Solution:
\[
x = \frac{5}{2}, \, x = -4
\]

---

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

#### Step 1: Factorize
We look for two numbers that multiply to \( -4 \) and add up to \( 3 \). These numbers are \( 4 \) and \( -1 \).

Rewrite the equation:
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]

#### Step 2: Solve for \( x \)
Set each factor equal to zero:
\[
x + 4 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = -4 \quad \text{or} \quad x = 1
\]

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

---

Equation 5: \( x^2 - 9 = 0 \)

#### Step 1: Recognize the difference of squares
\[
x^2 - 9 = (x - 3)(x + 3)
\]

#### Step 2: Solve for \( x \)
Set each factor equal to zero:
\[
x - 3 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = 3 \quad \text{or} \quad x = -3
\]

#### Solution:
\[
x = 3, \, x = -3
\]

---

Equation 6: \( x^2 - 6x - 16 = 0 \)

#### Step 1: Factorize
We look for two numbers that multiply to \( -16 \) and add up to \( -6 \). These numbers are \( -8 \) and \( 2 \).

Rewrite the equation:
\[
x^2 - 6x - 16 = (x - 8)(x + 2)
\]

#### Step 2: Solve for \( x \)
Set each factor equal to zero:
\[
x - 8 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 8 \quad \text{or} \quad x = -2
\]

#### Solution:
\[
x = 8, \, x = -2
\]

---

Equation 7: \( x^2 + 6x - 10 = 30 \)

#### Step 1: Simplify
Subtract 30 from both sides:
\[
x^2 + 6x - 10 - 30 = 0
\]
\[
x^2 + 6x - 40 = 0
\]

#### Step 2: Use the quadratic formula
Here, \( a = 1 \), \( b = 6 \), and \( c = -40 \).

Calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-40) = 36 + 160 = 196
\]

Take the square root of the discriminant:
\[
\sqrt{196} = 14
\]

Apply the quadratic formula:
\[
x = \frac{-6 \pm 14}{2 \cdot 1} = \frac{-6 \pm 14}{2}
\]

Solve for the two roots:
\[
x = \frac{-6 + 14}{2} = \frac{8}{2} = 4
\]
\[
x = \frac{-6 - 14}{2} = \frac{-20}{2} = -10
\]

#### Solution:
\[
x = 4, \, x = -10
\]

---

Equation 8: \( 4x^2 + 2x - 12 = 0 \)

#### Step 1: Simplify
Divide the entire equation by 2:
\[
2x^2 + x - 6 = 0
\]

#### Step 2: Factorize
We look for two numbers that multiply to \( 2 \times (-6) = -12 \) and add up to \( 1 \). These numbers are \( 4 \) and \( -3 \).

Rewrite the middle term:
\[
2x^2 + x - 6 = 2x^2 + 4x - 3x - 6
\]

Factor by grouping:
\[
= 2x(x + 2) - 3(x + 2)
\]
\[
= (2x - 3)(x + 2)
\]

#### Step 3: Solve for \( x \)
Set each factor equal to zero:
\[
2x - 3 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}
\]
\[
x = -2
\]

#### Solution:
\[
x = \frac{3}{2}, \, x = -2
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad x = \frac{3}{2}, \, x = \frac{4}{5} \\
2. & \quad x = \frac{7}{3}, \, x = -7 \\
3. & \quad x = \frac{5}{2}, \, x = -4 \\
4. & \quad x = -4, \, x = 1 \\
5. & \quad x = 3, \, x = -3 \\
6. & \quad x = 8, \, x = -2 \\
7. & \quad x = 4, \, x = -10 \\
8. & \quad x = \frac{3}{2}, \, x = -2 \\
\end{aligned}
}
\]