How to solve (-x-3)^(2)=8

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (-x-3)^(2)=8 | Equatix

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Question

$${ \left(-x-3 \right) }^{ 2 } =8$$

Solve for x

$x=-2\sqrt{2}-3\approx -5.828427125$
$x=2\sqrt{2}-3\approx -0.171572875$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$-x-3=2\sqrt{2}$$ $$-x-3=-2\sqrt{2}$$

Add $3$ to both sides of the equation.

$$-x-3-\left(-3\right)=2\sqrt{2}-\left(-3\right)$$ $$-x-3-\left(-3\right)=-2\sqrt{2}-\left(-3\right)$$

Subtracting $-3$ from itself leaves $0$.

$$-x=2\sqrt{2}-\left(-3\right)$$ $$-x=-2\sqrt{2}-\left(-3\right)$$

Subtract $-3$ from $2\sqrt{2}$.

$$-x=2\sqrt{2}+3$$

Subtract $-3$ from $-2\sqrt{2}$.

$$-x=3-2\sqrt{2}$$

Divide both sides by $-1$.

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

Dividing by $-1$ undoes the multiplication by $-1$.

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

Divide $2\sqrt{2}+3$ by $-1$.

$$x=-\left(2\sqrt{2}+3\right)$$

Divide $-2\sqrt{2}+3$ by $-1$.

$$x=2\sqrt{2}-3$$

The equation is now solved.

$$x=-\left(2\sqrt{2}+3\right)$$ $$x=2\sqrt{2}-3$$

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