Solve -left(3-xright)^2+1=-5

$$- { \left(3-x \right) }^{ 2 } +1=-5$$

Solve for x

$$x=3-\sqrt{6}\approx 0.550510257$$

$$x=\sqrt{6}+3\approx 5.449489743$$

Steps by Finding Square Root

Subtract 1 from both sides of the equation.

$$-\left(-x+3\right)^{2}+1-1=-5-1$$

Subtracting 1 from itself leaves 0.

$$-\left(-x+3\right)^{2}=-5-1$$

Subtract 1 from -5.

$$-\left(-x+3\right)^{2}=-6$$

Divide both sides by -1.

$$\frac{-\left(-x+3\right)^{2}}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\left(-x+3\right)^{2}=-\frac{6}{-1}$$

Divide -6 by -1.

$$\left(-x+3\right)^{2}=6$$

Take the square root of both sides of the equation.

$$-x+3=\sqrt{6} -x+3=-\sqrt{6}$$

Subtract 3 from both sides of the equation.

$$-x+3-3=\sqrt{6}-3 -x+3-3=-\sqrt{6}-3$$

Subtracting 3 from itself leaves 0.

$$-x=\sqrt{6}-3 -x=-\sqrt{6}-3$$

Subtract 3 from \sqrt{6}.

$$-x=\sqrt{6}-3$$

Subtract 3 from -\sqrt{6}.

$$-x=-\sqrt{6}-3$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\sqrt{6}-3}{-1} \frac{-x}{-1}=\frac{-\sqrt{6}-3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\sqrt{6}-3}{-1} x=\frac{-\sqrt{6}-3}{-1}$$

Divide \sqrt{6}-3 by -1.

$$x=3-\sqrt{6}$$

Divide -\sqrt{6}-3 by -1.

$$x=\sqrt{6}+3$$

The equation is now solved.

$$x=3-\sqrt{6} x=\sqrt{6}+3$$