To solve the quadratic equation:

\[
2x^{2} + 3x - 5 = 0
\]

we can use the **quadratic formula**, which is given by:

\[
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\]

where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^{2} + bx + c = 0 \).

For the given equation:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -5 \)

**Step 1: Calculate the Discriminant (\( \Delta \))**

\[
\Delta = b^{2} - 4ac = (3)^{2} - 4 \times 2 \times (-5) = 9 + 40 = 49
\]

**Step 2: Compute the Square Root of the Discriminant**

\[
\sqrt{\Delta} = \sqrt{49} = 7
\]

**Step 3: Apply the Quadratic Formula**

\[
x = \frac{-3 \pm 7}{2 \times 2} = \frac{-3 \pm 7}{4}
\]

This yields two solutions:

1. **First Solution (\( + \) sign):**
   \[
   x = \frac{-3 + 7}{4} = \frac{4}{4} = 1
   \]

2. **Second Solution (\( - \) sign):**
   \[
   x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}
   \]

**Conclusion:**

The solutions to the equation \( 2x^{2} + 3x - 5 = 0 \) are:

\[
x = 1 \quad \text{and} \quad x = -\frac{5}{2}
\]

Home Question Bank Math Algebra

Question

7. \( 2 x^{2}+3 x-5=0 \)

Ask by French Gardner. in the United States

Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions are \( x = 1 \) and \( x = -\frac{5}{2} \).

Solution

To solve the quadratic equation: \[ 2x^{2} + 3x - 5 = 0 \] we can use the **quadratic formula**, which is given by: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^{2} + bx + c = 0 \). For the given equation: - \( a = 2 \) - \( b = 3 \) - \( c = -5 \) **Step 1: Calculate the Discriminant (\( \Delta \))** \[ \Delta = b^{2} - 4ac = (3)^{2} - 4 \times 2 \times (-5) = 9 + 40 = 49 \] **Step 2: Compute the Square Root of the Discriminant** \[ \sqrt{\Delta} = \sqrt{49} = 7 \] **Step 3: Apply the Quadratic Formula** \[ x = \frac{-3 \pm 7}{2 \times 2} = \frac{-3 \pm 7}{4} \] This yields two solutions: 1. **First Solution (\( + \) sign):** \[ x = \frac{-3 + 7}{4} = \frac{4}{4} = 1 \] 2. **Second Solution (\( - \) sign):** \[ x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2} \] **Conclusion:** The solutions to the equation \( 2x^{2} + 3x - 5 = 0 \) are: \[ x = 1 \quad \text{and} \quad x = -\frac{5}{2} \]

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Bonus Knowledge

To solve the quadratic equation \( 2x^2 + 3x - 5 = 0 \), you can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 3 \), and \( c = -5 \). Plugging in these values gives you \( x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4} \). This leads to two potential solutions: \( x = 1 \) and \( x = -2.5 \). Fun fact: Quadratic equations like this one were studied by ancient civilizations, including the Babylonians. They used a geometric approach to solve these challenges long before modern algebra came into play! Their methods laid the groundwork for what we use today, which is just a mathematical treasure trove waiting to be explored! If you're solving similar equations, a common mistake is forgetting to simplify the discriminant (the part under the square root) correctly. Always check your sign and arithmetic! Mistakes like this can lead to errors in the provided solutions. Breaking down the equation step-by-step and double-checking your work can save you from unnecessary headaches. Happy solving!

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