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

$$(x-2) \times (x-3) \times (x-4)=(x-2) \times (x-3) \times (x-5)$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using the Quadratic Formula

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

$$\left(x^{2}-5x+6\right)\left(x-4\right)=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-5x+6 by x-4 and combine like terms.

$$x^{3}-9x^{2}+26x-24=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

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

$$x^{3}-9x^{2}+26x-24=\left(x^{2}-5x+6\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-5x+6 by x-5 and combine like terms.

$$x^{3}-9x^{2}+26x-24=x^{3}-10x^{2}+31x-30$$

Subtract x^{3} from both sides.

$$x^{3}-9x^{2}+26x-24-x^{3}=-10x^{2}+31x-30$$

Combine x^{3} and -x^{3} to get 0.

$$-9x^{2}+26x-24=-10x^{2}+31x-30$$

Add 10x^{2} to both sides.

$$-9x^{2}+26x-24+10x^{2}=31x-30$$

Combine -9x^{2} and 10x^{2} to get x^{2}.

$$x^{2}+26x-24=31x-30$$

Subtract 31x from both sides.

$$x^{2}+26x-24-31x=-30$$

Combine 26x and -31x to get -5x.

$$x^{2}-5x-24=-30$$

Add 30 to both sides.

$$x^{2}-5x-24+30=0$$

Add -24 and 30 to get 6.

$$x^{2}-5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±1}{2}$$

Now solve the equation x=\frac{5±1}{2} when ± is plus. Add 5 to 1.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=3 x=2$$

Steps for Completing the Square

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

$$\left(x^{2}-5x+6\right)\left(x-4\right)=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-5x+6 by x-4 and combine like terms.

$$x^{3}-9x^{2}+26x-24=\left(x-2\right)\left(x-3\right)\left(x-5\right)$$

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

$$x^{3}-9x^{2}+26x-24=\left(x^{2}-5x+6\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-5x+6 by x-5 and combine like terms.

$$x^{3}-9x^{2}+26x-24=x^{3}-10x^{2}+31x-30$$

Subtract x^{3} from both sides.

$$x^{3}-9x^{2}+26x-24-x^{3}=-10x^{2}+31x-30$$

Combine x^{3} and -x^{3} to get 0.

$$-9x^{2}+26x-24=-10x^{2}+31x-30$$

Add 10x^{2} to both sides.

$$-9x^{2}+26x-24+10x^{2}=31x-30$$

Combine -9x^{2} and 10x^{2} to get x^{2}.

$$x^{2}+26x-24=31x-30$$

Subtract 31x from both sides.

$$x^{2}+26x-24-31x=-30$$

Combine 26x and -31x to get -5x.

$$x^{2}-5x-24=-30$$

Add 24 to both sides.

$$x^{2}-5x=-30+24$$

Add -30 and 24 to get -6.

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

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

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

Factor x^{2}-5x+\frac{25}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$