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

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

Solve for x

$$x=1$$

$$x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4-2x by 3-x and combine like terms.

$$12-10x+2x^{2}=4$$

Subtract 4 from both sides.

$$12-10x+2x^{2}-4=0$$

Subtract 4 from 12 to get 8.

$$8-10x+2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-10x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times 8}}{2\times 2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times 8}}{2\times 2}$$

Multiply -4 times 2.

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

Multiply -8 times 8.

$$x=\frac{-\left(-10\right)±\sqrt{100-64}}{2\times 2}$$

Add 100 to -64.

$$x=\frac{-\left(-10\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$x=\frac{-\left(-10\right)±6}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10±6}{4}$$

Now solve the equation x=\frac{10±6}{4} when ± is plus. Add 10 to 6.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{10±6}{4} when ± is minus. Subtract 6 from 10.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=4 x=1$$

Steps for Completing the Square

Use the distributive property to multiply 4-2x by 3-x and combine like terms.

$$12-10x+2x^{2}=4$$

Subtract 12 from both sides.

$$-10x+2x^{2}=4-12$$

Subtract 12 from 4 to get -8.

$$-10x+2x^{2}=-8$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-10x=-8$$

Divide both sides by 2.

$$\frac{2x^{2}-10x}{2}=-\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=-\frac{8}{2}$$

Divide -10 by 2.

$$x^{2}-5x=-\frac{8}{2}$$

Divide -8 by 2.

$$x^{2}-5x=-4$$

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.

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

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

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

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

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

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

$$x=4 x=1$$