Solve (2x-5)(x-4)-7=(x-2)(x-3)

$$( 2 x - 5 ) ( x - 4 ) - 7 = ( x - 2 ) ( x - 3 )$$

Solve for x

$$x=1$$

$$x=7$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-5 by x-4 and combine like terms.

$$2x^{2}-13x+20-7=\left(x-2\right)\left(x-3\right)$$

Subtract 7 from 20 to get 13.

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

Use the distributive property to multiply x-2 by x-3 and combine like terms.

$$2x^{2}-13x+13=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$2x^{2}-13x+13-x^{2}=-5x+6$$

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

$$x^{2}-13x+13=-5x+6$$

Add 5x to both sides.

$$x^{2}-13x+13+5x=6$$

Combine -13x and 5x to get -8x.

$$x^{2}-8x+13=6$$

Subtract 6 from both sides.

$$x^{2}-8x+13-6=0$$

Subtract 6 from 13 to get 7.

$$x^{2}-8x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 7}}{2}$$

Multiply -4 times 7.

$$x=\frac{-\left(-8\right)±\sqrt{64-28}}{2}$$

Add 64 to -28.

$$x=\frac{-\left(-8\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-8\right)±6}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±6}{2}$$

Now solve the equation x=\frac{8±6}{2} when ± is plus. Add 8 to 6.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{8±6}{2} when ± is minus. Subtract 6 from 8.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=7 x=1$$

Steps for Completing the Square

Use the distributive property to multiply 2x-5 by x-4 and combine like terms.

$$2x^{2}-13x+20-7=\left(x-2\right)\left(x-3\right)$$

Subtract 7 from 20 to get 13.

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

Use the distributive property to multiply x-2 by x-3 and combine like terms.

$$2x^{2}-13x+13=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$2x^{2}-13x+13-x^{2}=-5x+6$$

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

$$x^{2}-13x+13=-5x+6$$

Add 5x to both sides.

$$x^{2}-13x+13+5x=6$$

Combine -13x and 5x to get -8x.

$$x^{2}-8x+13=6$$

Subtract 13 from both sides.

$$x^{2}-8x=6-13$$

Subtract 13 from 6 to get -7.

$$x^{2}-8x=-7$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-7+16$$

Add -7 to 16.

$$x^{2}-8x+16=9$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=3 x-4=-3$$

Add 4 to both sides of the equation.

$$x=7 x=1$$