Solve (x-3)(x-4)=6x

$$( x - 3 ) ( x - 4 ) = 6 x$$

Solve for x

$$x=1$$

$$x=12$$

Steps Using Factoring

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$x^{2}-7x+12=6x$$

Subtract 6x from both sides.

$$x^{2}-7x+12-6x=0$$

Combine -7x and -6x to get -13x.

$$x^{2}-13x+12=0$$

To solve the equation, factor x^{2}-13x+12 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-12\right)\left(x-1\right)$$

To find equation solutions, solve x-12=0 and x-1=0.

$$x=12 x=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$x^{2}-7x+12=6x$$

Subtract 6x from both sides.

$$x^{2}-7x+12-6x=0$$

Combine -7x and -6x to get -13x.

$$x^{2}-13x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite x^{2}-13x+12 as \left(x^{2}-12x\right)+\left(-x+12\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-12\right)-\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x-1\right)$$

To find equation solutions, solve x-12=0 and x-1=0.

$$x=12 x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$x^{2}-7x+12=6x$$

Subtract 6x from both sides.

$$x^{2}-7x+12-6x=0$$

Combine -7x and -6x to get -13x.

$$x^{2}-13x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 12}}{2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-13\right)±\sqrt{169-48}}{2}$$

Add 169 to -48.

$$x=\frac{-\left(-13\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-\left(-13\right)±11}{2}$$

The opposite of -13 is 13.

$$x=\frac{13±11}{2}$$

Now solve the equation x=\frac{13±11}{2} when ± is plus. Add 13 to 11.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{13±11}{2} when ± is minus. Subtract 11 from 13.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=12 x=1$$

Steps for Completing the Square

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$x^{2}-7x+12=6x$$

Subtract 6x from both sides.

$$x^{2}-7x+12-6x=0$$

Combine -7x and -6x to get -13x.

$$x^{2}-13x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-13x=-12$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-12+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=-12+\frac{169}{4}$$

Add -12 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{121}{4}$$

Factor x^{2}-13x+\frac{169}{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}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{11}{2} x-\frac{13}{2}=-\frac{11}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$x=12 x=1$$