Solve 4=(3x-2)^2-9

$$4 = ( 3 x - 2 ) ^ { 2 } - 9$$

Solve for x

$$x = \frac{\sqrt{13} + 2}{3} \approx 1.868517092$$

$$x=\frac{2-\sqrt{13}}{3}\approx -0.535183758$$

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}.

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

Subtract 9 from 4 to get -5.

$$4=9x^{2}-12x-5$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}-12x-5=4$$

Subtract 4 from both sides.

$$9x^{2}-12x-5-4=0$$

Subtract 4 from -5 to get -9.

$$9x^{2}-12x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-9\right)}}{2\times 9}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$x=\frac{-\left(-12\right)±\sqrt{144+324}}{2\times 9}$$

Add 144 to 324.

$$x=\frac{-\left(-12\right)±\sqrt{468}}{2\times 9}$$

Take the square root of 468.

$$x=\frac{-\left(-12\right)±6\sqrt{13}}{2\times 9}$$

The opposite of -12 is 12.

$$x=\frac{12±6\sqrt{13}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{12±6\sqrt{13}}{18}$$

Now solve the equation x=\frac{12±6\sqrt{13}}{18} when ± is plus. Add 12 to 6\sqrt{13}.

$$x=\frac{6\sqrt{13}+12}{18}$$

Divide 12+6\sqrt{13} by 18.

$$x=\frac{\sqrt{13}+2}{3}$$

Now solve the equation x=\frac{12±6\sqrt{13}}{18} when ± is minus. Subtract 6\sqrt{13} from 12.

$$x=\frac{12-6\sqrt{13}}{18}$$

Divide 12-6\sqrt{13} by 18.

$$x=\frac{2-\sqrt{13}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+2}{3} x=\frac{2-\sqrt{13}}{3}$$

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}.

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

Subtract 9 from 4 to get -5.

$$4=9x^{2}-12x-5$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}-12x-5=4$$

Add 5 to both sides.

$$9x^{2}-12x=4+5$$

Add 4 and 5 to get 9.

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

Divide both sides by 9.

$$\frac{9x^{2}-12x}{9}=\frac{9}{9}$$

Dividing by 9 undoes the multiplication by 9.

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

Reduce the fraction \frac{-12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{4}{3}x=\frac{9}{9}$$

Divide 9 by 9.

$$x^{2}-\frac{4}{3}x=1$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=1+\frac{4}{9}$$

Add 1 to \frac{4}{9}.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{13}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{13}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{13}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{13}}{3} x-\frac{2}{3}=-\frac{\sqrt{13}}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{\sqrt{13}+2}{3} x=\frac{2-\sqrt{13}}{3}$$