Solve (3x+2)(x+2)-4(x+1)^2=0

$$( 3 x + 2 ) ( x + 2 ) - 4 ( x + 1 ) ^ { 2 } = 0$$

Solve for x

$$x=0$$

Steps by Finding Square Root

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

$$3x^{2}+8x+4-4\left(x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$3x^{2}+8x+4-4\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$3x^{2}+8x+4-4x^{2}-8x-4=0$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}+8x+4-8x-4=0$$

Combine 8x and -8x to get 0.

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

Subtract 4 from 4 to get 0.

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

Divide both sides by -1. 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

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

$$3x^{2}+8x+4-4\left(x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$3x^{2}+8x+4-4\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$3x^{2}+8x+4-4x^{2}-8x-4=0$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}+8x+4-8x-4=0$$

Combine 8x and -8x to get 0.

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

Subtract 4 from 4 to get 0.

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

Divide both sides by -1. 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.

$$x=0$$