What are the solutions to the equation \( (x-1)(x+2)=-2 \) ? \[ \begin{array}{l}x=-3, x=0 \\ x=-1, x=3 \\ x=-2, x=1 \\ x=0\end{array} \]

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(\left(x-1\right)\left(x+2\right)+2=0\) - step1: Expand the expression: \(x^{2}+x=0\) - step2: Factor the expression: \(x\left(x+1\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x=0\\&x+1=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=0\\&x=-1\end{align}\) - step5: Rewrite: \(x_{1}=-1,x_{2}=0\) To solve the equation \( (x-1)(x+2) = -2 \), we first rearrange it to standard form: \[ (x-1)(x+2) + 2 = 0 \] Now, we can expand the left side: \[ x^2 + 2x - x - 2 + 2 = 0 \] This simplifies to: \[ x^2 + x = 0 \] Next, we can factor this equation: \[ x(x + 1) = 0 \] Setting each factor to zero gives us the solutions: 1. \( x = 0 \) 2. \( x + 1 = 0 \) which leads to \( x = -1 \) Thus, the solutions to the equation \( (x-1)(x+2) = -2 \) are: \[ x = -1 \quad \text{and} \quad x = 0 \] From the provided options, the correct pair of solutions is \( x = -1, x = 0 \).