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

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

Solve for x

$$x = \frac{16}{9} = 1\frac{7}{9} \approx 1.777777778$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

$$9x^{2}-12x+4=4x+4$$

Subtract 4x from both sides.

$$9x^{2}-12x+4-4x=4$$

Combine -12x and -4x to get -16x.

$$9x^{2}-16x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-16x+4-4=0$$

Subtract 4 from 4 to get 0.

$$9x^{2}-16x=0$$

Factor out x.

$$x\left(9x-16\right)=0$$

To find equation solutions, solve x=0 and 9x-16=0.

$$x=0 x=\frac{16}{9}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

$$9x^{2}-12x+4=4x+4$$

Subtract 4x from both sides.

$$9x^{2}-12x+4-4x=4$$

Combine -12x and -4x to get -16x.

$$9x^{2}-16x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-16x+4-4=0$$

Subtract 4 from 4 to get 0.

$$9x^{2}-16x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -16 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}}}{2\times 9}$$

Take the square root of \left(-16\right)^{2}.

$$x=\frac{-\left(-16\right)±16}{2\times 9}$$

The opposite of -16 is 16.

$$x=\frac{16±16}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{16±16}{18}$$

Now solve the equation x=\frac{16±16}{18} when ± is plus. Add 16 to 16.

$$x=\frac{32}{18}$$

Reduce the fraction \frac{32}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{16}{9}$$

Now solve the equation x=\frac{16±16}{18} when ± is minus. Subtract 16 from 16.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{16}{9} x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

$$9x^{2}-12x+4=4x+4$$

Subtract 4x from both sides.

$$9x^{2}-12x+4-4x=4$$

Combine -12x and -4x to get -16x.

$$9x^{2}-16x+4=4$$

Subtract 4 from both sides.

$$9x^{2}-16x=4-4$$

Subtract 4 from 4 to get 0.

$$9x^{2}-16x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-16x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{16}{9}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{16}{9}x=0$$

Divide -\frac{16}{9}, the coefficient of the x term, by 2 to get -\frac{8}{9}. Then add the square of -\frac{8}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{9}x+\left(-\frac{8}{9}\right)^{2}=\left(-\frac{8}{9}\right)^{2}$$

Square -\frac{8}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{9}x+\frac{64}{81}=\frac{64}{81}$$

Factor x^{2}-\frac{16}{9}x+\frac{64}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{8}{9}\right)^{2}=\frac{64}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{9}\right)^{2}}=\sqrt{\frac{64}{81}}$$

Simplify.

$$x-\frac{8}{9}=\frac{8}{9} x-\frac{8}{9}=-\frac{8}{9}$$

Add \frac{8}{9} to both sides of the equation.

$$x=\frac{16}{9} x=0$$