Solve left(-4x-3right)^2=2

$${ \left(-4x-3 \right) }^{ 2 } =2$$

Solve for x

$$x=\frac{-\sqrt{2}-3}{4}\approx -1.103553391$$

$$x=\frac{\sqrt{2}-3}{4}\approx -0.396446609$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$-4x-3=\sqrt{2} -4x-3=-\sqrt{2}$$

Add 3 to both sides of the equation.

$$-4x-3-\left(-3\right)=\sqrt{2}-\left(-3\right) -4x-3-\left(-3\right)=-\sqrt{2}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$-4x=\sqrt{2}-\left(-3\right) -4x=-\sqrt{2}-\left(-3\right)$$

Subtract -3 from \sqrt{2}.

$$-4x=\sqrt{2}+3$$

Subtract -3 from -\sqrt{2}.

$$-4x=3-\sqrt{2}$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{\sqrt{2}+3}{-4} \frac{-4x}{-4}=\frac{3-\sqrt{2}}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{\sqrt{2}+3}{-4} x=\frac{3-\sqrt{2}}{-4}$$

Divide \sqrt{2}+3 by -4.

$$x=\frac{-\sqrt{2}-3}{4}$$

Divide -\sqrt{2}+3 by -4.

$$x=\frac{\sqrt{2}-3}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{2}-3}{4} x=\frac{\sqrt{2}-3}{4}$$