Solve (3x+1)(3x-1)+(x-2)^2/2=1-2x

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

Solve for x

$$x=0$$

Steps by Finding Square Root

Multiply both sides of the equation by 2.

$$2\left(3x+1\right)\left(3x-1\right)+\left(x-2\right)^{2}=2-4x$$

Use the distributive property to multiply 2 by 3x+1.

$$\left(6x+2\right)\left(3x-1\right)+\left(x-2\right)^{2}=2-4x$$

Use the distributive property to multiply 6x+2 by 3x-1 and combine like terms.

$$18x^{2}-2+\left(x-2\right)^{2}=2-4x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$18x^{2}-2+x^{2}-4x+4=2-4x$$

Combine 18x^{2} and x^{2} to get 19x^{2}.

$$19x^{2}-2-4x+4=2-4x$$

Add -2 and 4 to get 2.

$$19x^{2}+2-4x=2-4x$$

Add 4x to both sides.

$$19x^{2}+2-4x+4x=2$$

Combine -4x and 4x to get 0.

$$19x^{2}+2=2$$

Subtract 2 from both sides.

$$19x^{2}=2-2$$

Subtract 2 from 2 to get 0.

$$19x^{2}=0$$

Divide both sides by 19. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2\left(3x+1\right)\left(3x-1\right)+\left(x-2\right)^{2}=2-4x$$

Use the distributive property to multiply 2 by 3x+1.

$$\left(6x+2\right)\left(3x-1\right)+\left(x-2\right)^{2}=2-4x$$

Use the distributive property to multiply 6x+2 by 3x-1 and combine like terms.

$$18x^{2}-2+\left(x-2\right)^{2}=2-4x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$18x^{2}-2+x^{2}-4x+4=2-4x$$

Combine 18x^{2} and x^{2} to get 19x^{2}.

$$19x^{2}-2-4x+4=2-4x$$

Add -2 and 4 to get 2.

$$19x^{2}+2-4x=2-4x$$

Subtract 2 from both sides.

$$19x^{2}+2-4x-2=-4x$$

Subtract 2 from 2 to get 0.

$$19x^{2}-4x=-4x$$

Add 4x to both sides.

$$19x^{2}-4x+4x=0$$

Combine -4x and 4x to get 0.

$$19x^{2}=0$$

Divide both sides by 19. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$