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

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

Solve for x

$$x = \frac{\sqrt{21} + 7}{2} \approx 5.791287847$$

$$x = \frac{7 - \sqrt{21}}{2} \approx 1.208712153$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply -3 by x-1.

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

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4+3=0$$

Add 4 and 3 to get 7.

$$x^{2}-7x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 7}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 7}}{2}$$

Multiply -4 times 7.

$$x=\frac{-\left(-7\right)±\sqrt{49-28}}{2}$$

Add 49 to -28.

$$x=\frac{-\left(-7\right)±\sqrt{21}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{21}}{2}$$

Now solve the equation x=\frac{7±\sqrt{21}}{2} when ± is plus. Add 7 to \sqrt{21}.

$$x=\frac{\sqrt{21}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{21}}{2} when ± is minus. Subtract \sqrt{21} from 7.

$$x=\frac{7-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+7}{2} x=\frac{7-\sqrt{21}}{2}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply -3 by x-1.

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

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4+3=0$$

Add 4 and 3 to get 7.

$$x^{2}-7x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-7x=-7$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-7+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-7+\frac{49}{4}$$

Add -7 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{21}{4}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{21}}{2} x-\frac{7}{2}=-\frac{\sqrt{21}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}+7}{2} x=\frac{7-\sqrt{21}}{2}$$