Solve (2x-5)(x-2)=left(x-1right)^2+4

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

Solve for x

$$x = \frac{\sqrt{29} + 7}{2} \approx 6.192582404$$

$$x=\frac{7-\sqrt{29}}{2}\approx 0.807417596$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-5 by x-2 and combine like terms.

$$2x^{2}-9x+10=\left(x-1\right)^{2}+4$$

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

$$2x^{2}-9x+10=x^{2}-2x+1+4$$

Add 1 and 4 to get 5.

$$2x^{2}-9x+10=x^{2}-2x+5$$

Subtract x^{2} from both sides.

$$2x^{2}-9x+10-x^{2}=-2x+5$$

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

$$x^{2}-9x+10=-2x+5$$

Add 2x to both sides.

$$x^{2}-9x+10+2x=5$$

Combine -9x and 2x to get -7x.

$$x^{2}-7x+10=5$$

Subtract 5 from both sides.

$$x^{2}-7x+10-5=0$$

Subtract 5 from 10 to get 5.

$$x^{2}-7x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-7\right)±\sqrt{49-20}}{2}$$

Add 49 to -20.

$$x=\frac{-\left(-7\right)±\sqrt{29}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{29}}{2}$$

Now solve the equation x=\frac{7±\sqrt{29}}{2} when ± is plus. Add 7 to \sqrt{29}.

$$x=\frac{\sqrt{29}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{29}}{2} when ± is minus. Subtract \sqrt{29} from 7.

$$x=\frac{7-\sqrt{29}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{29}+7}{2} x=\frac{7-\sqrt{29}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2x-5 by x-2 and combine like terms.

$$2x^{2}-9x+10=\left(x-1\right)^{2}+4$$

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

$$2x^{2}-9x+10=x^{2}-2x+1+4$$

Add 1 and 4 to get 5.

$$2x^{2}-9x+10=x^{2}-2x+5$$

Subtract x^{2} from both sides.

$$2x^{2}-9x+10-x^{2}=-2x+5$$

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

$$x^{2}-9x+10=-2x+5$$

Add 2x to both sides.

$$x^{2}-9x+10+2x=5$$

Combine -9x and 2x to get -7x.

$$x^{2}-7x+10=5$$

Subtract 10 from both sides.

$$x^{2}-7x=5-10$$

Subtract 10 from 5 to get -5.

$$x^{2}-7x=-5$$

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.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-5+\left(-\frac{7}{2}\right)^{2}$$

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

$$x^{2}-7x+\frac{49}{4}=-5+\frac{49}{4}$$

Add -5 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{29}{4}$$

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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{29}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{29}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{29}}{2} x-\frac{7}{2}=-\frac{\sqrt{29}}{2}$$

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

$$x=\frac{\sqrt{29}+7}{2} x=\frac{7-\sqrt{29}}{2}$$