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

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

Solve for x

$$x=\frac{\sqrt{29}-2}{5}\approx 0.677032961$$

$$x=\frac{-\sqrt{29}-2}{5}\approx -1.477032961$$

Steps Using the Quadratic Formula

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

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

Consider \left(3x-2\right)\left(3x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Expand \left(3x\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-4x+1=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=-4$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1=-4$$

Add 4 to both sides.

$$-5x^{2}-4x+1+4=0$$

Add 1 and 4 to get 5.

$$-5x^{2}-4x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -4 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16+100}}{2\left(-5\right)}$$

Add 16 to 100.

$$x=\frac{-\left(-4\right)±\sqrt{116}}{2\left(-5\right)}$$

Take the square root of 116.

$$x=\frac{-\left(-4\right)±2\sqrt{29}}{2\left(-5\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{29}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{4±2\sqrt{29}}{-10}$$

Now solve the equation x=\frac{4±2\sqrt{29}}{-10} when ± is plus. Add 4 to 2\sqrt{29}.

$$x=\frac{2\sqrt{29}+4}{-10}$$

Divide 4+2\sqrt{29} by -10.

$$x=\frac{-\sqrt{29}-2}{5}$$

Now solve the equation x=\frac{4±2\sqrt{29}}{-10} when ± is minus. Subtract 2\sqrt{29} from 4.

$$x=\frac{4-2\sqrt{29}}{-10}$$

Divide 4-2\sqrt{29} by -10.

$$x=\frac{\sqrt{29}-2}{5}$$

The equation is now solved.

$$x=\frac{-\sqrt{29}-2}{5} x=\frac{\sqrt{29}-2}{5}$$

Steps for Completing the Square

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

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

Consider \left(3x-2\right)\left(3x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Expand \left(3x\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-4x+1=9x^{2}-4$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=-4$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1=-4$$

Subtract 1 from both sides.

$$-5x^{2}-4x=-4-1$$

Subtract 1 from -4 to get -5.

$$-5x^{2}-4x=-5$$

Divide both sides by -5.

$$\frac{-5x^{2}-4x}{-5}=-\frac{5}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{4}{-5}\right)x=-\frac{5}{-5}$$

Divide -4 by -5.

$$x^{2}+\frac{4}{5}x=-\frac{5}{-5}$$

Divide -5 by -5.

$$x^{2}+\frac{4}{5}x=1$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=1+\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=1+\frac{4}{25}$$

Add 1 to \frac{4}{25}.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{29}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{29}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{29}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{\sqrt{29}}{5} x+\frac{2}{5}=-\frac{\sqrt{29}}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=\frac{\sqrt{29}-2}{5} x=\frac{-\sqrt{29}-2}{5}$$