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

To solve the quadratic equation \( 3x^{2} + 2x - 5 = 0 \), you can use either **factoring** or the **quadratic formula**. Below are both methods: --- ### **Method 1: Factoring** 1. **Write the equation in standard form:** \[ 3x^{2} + 2x - 5 = 0 \] 2. **Find two numbers that multiply to \( a \times c = 3 \times (-5) = -15 \) and add to \( b = 2 \):** The numbers are **5** and **-3** because: \[ 5 \times (-3) = -15 \quad \text{and} \quad 5 + (-3) = 2 \] 3. **Rewrite the middle term using these numbers:** \[ 3x^{2} + 5x - 3x - 5 = 0 \] 4. **Factor by grouping:** \[ (3x^{2} + 5x) - (3x + 5) = 0 \\ x(3x + 5) -1(3x + 5) = 0 \] 5. **Factor out the common binomial factor:** \[ (3x + 5)(x - 1) = 0 \] 6. **Set each factor equal to zero and solve for \( x \):** \[ 3x + 5 = 0 \quad \Rightarrow \quad x = -\frac{5}{3} \\ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] **Solutions:** \[ x = 1 \quad \text{and} \quad x = -\frac{5}{3} \] --- ### **Method 2: Quadratic Formula** For any quadratic equation of the form \( ax^{2} + bx + c = 0 \), the solutions are given by: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] 1. **Identify the coefficients:** \[ a = 3, \quad b = 2, \quad c = -5 \] 2. **Plug the coefficients into the quadratic formula:** \[ x = \frac{-2 \pm \sqrt{2^{2} - 4 \times 3 \times (-5)}}{2 \times 3} \] 3. **Calculate the discriminant (\( \Delta \)):** \[ \Delta = 2^{2} - 4 \times 3 \times (-5) = 4 + 60 = 64 \] 4. **Take the square root of the discriminant:** \[ \sqrt{64} = 8 \] 5. **Find the two solutions:** \[ x = \frac{-2 + 8}{6} = \frac{6}{6} = 1 \\ x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3} \] **Solutions:** \[ x = 1 \quad \text{and} \quad x = -\frac{5}{3} \] --- Both methods yield the same solutions: \[ \boxed{x = 1 \quad \text{and} \quad x = -\dfrac{5}{3}} \]