Solve 2x(x-3)-4x+12=0

$$2 x ( x - 3 ) - 4 x + 12 = 0$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x by x-3.

$$2x^{2}-6x-4x+12=0$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+12=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-3\right)-2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-2\right)$$

To find equation solutions, solve x-3=0 and x-2=0.

$$x=3 x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-3.

$$2x^{2}-6x-4x+12=0$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times 12}}{2\times 2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 2}$$

Add 100 to -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$x=\frac{-\left(-10\right)±2}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10±2}{4}$$

Now solve the equation x=\frac{10±2}{4} when ± is plus. Add 10 to 2.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{10±2}{4} when ± is minus. Subtract 2 from 10.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

The equation is now solved.

$$x=3 x=2$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-3.

$$2x^{2}-6x-4x+12=0$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+12=0$$

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

$$2x^{2}-10x=-12$$

Divide both sides by 2.

$$\frac{2x^{2}-10x}{2}=-\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=-\frac{12}{2}$$

Divide -10 by 2.

$$x^{2}-5x=-\frac{12}{2}$$

Divide -12 by 2.

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

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.

$$x=3 x=2$$