4. \( y=2 x^{2}+3 x-20 \)

**Step 1. Setting the quadratic equal to zero** We start with the equation \[ y = 2x^2 + 3x - 20 \] and find the zeros by setting \( y = 0 \): \[ 2x^2 + 3x - 20 = 0 \] **Step 2. Factoring the quadratic** We need to factor the quadratic expression. We look for two numbers that multiply to \( 2 \times (-20) = -40 \) and add up to \( 3 \). The numbers \( 8 \) and \( -5 \) satisfy: \[ 8 \times (-5) = -40 \quad \text{and} \quad 8 + (-5) = 3 \] Rewrite the middle term \( 3x \) using \( 8x - 5x \): \[ 2x^2 + 8x - 5x - 20 = 0 \] Now, factor by grouping: - Group the first two terms and the last two terms: \[ (2x^2 + 8x) + (-5x - 20) = 0 \] - Factor out the greatest common factor (GCF) from each group: \[ 2x(x + 4) - 5(x + 4) = 0 \] - Notice that \( (x+4) \) is a common factor: \[ (x+4)(2x-5) = 0 \] **Step 3. Solving each factor** Set each factor equal to zero: 1. For \( x + 4 = 0 \): \[ x = -4 \] 2. For \( 2x - 5 = 0 \): \[ 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2} \] **Final Answer** The solutions are: \[ x = -4 \quad \text{or} \quad x = \frac{5}{2} \]