Solve the Quadratic Equation: Finding x in -x² + x - 2 = -2x^2 - 2x - 4

Solve the following equation:

$-x^2+x-2=-2x^2-2x-4$

To solve the equation $-x^2 + x - 2 = -2x^2 - 2x - 4$, we will proceed with the following steps:

• Step 1: Simplify the equation by moving all terms to one side of the equation.
• Step 2: Combine like terms to form a quadratic equation.
• Step 3: Use the quadratic formula to solve for $x$.

Now, let's go through these steps:

Step 1: Start with the given equation:

$-x^2 + x - 2 = -2x^2 - 2x - 4$

Add $2x^2$, $2x$, and $4$ to both sides to move all terms to the left side:

$-x^2 + x - 2 + 2x^2 + 2x + 4 = 0$

Step 2: Combine like terms:

$(2x^2 - x^2) + (x + 2x) + (-2 + 4) = 0$

This simplifies to:

$x^2 + 3x + 2 = 0$

Step 3: Use the quadratic formula $(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a})$ with $a = 1$, $b = 3$, and $c = 2$.

Calculate the discriminant: $b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1$.

Since the discriminant is positive, there are two distinct real roots. Substitute into the quadratic formula:

$x = \frac{{-3 \pm \sqrt{1}}}{2 \cdot 1} = \frac{{-3 \pm 1}}{2}$

Calculate the roots:

$x_1 = \frac{{-3 + 1}}{2} = \frac{{-2}}{2} = -1$

$x_2 = \frac{{-3 - 1}}{2} = \frac{{-4}}{2} = -2$

Therefore, the solutions to the equation are $x_1 = -2$ and $x_2 = -1$.