Solve (1-x)*(4-x)-4=0

$$( 1 - x ) \cdot ( 4 - x ) - 4 = 0$$

Solve for x

$$x=5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 1-x by 4-x and combine like terms.

$$4-5x+x^{2}-4=0$$

Subtract 4 from 4 to get 0.

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

Factor out x.

$$x\left(-5+x\right)=0$$

To find equation solutions, solve x=0 and -5+x=0.

$$x=0 x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1-x by 4-x and combine like terms.

$$4-5x+x^{2}-4=0$$

Subtract 4 from 4 to get 0.

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

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.

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

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

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

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2}$$

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

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

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

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=5 x=0$$

Steps for Completing the Square

Use the distributive property to multiply 1-x by 4-x and combine like terms.

$$4-5x+x^{2}-4=0$$

Subtract 4 from 4 to get 0.

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

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.

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

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.

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

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

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

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=5 x=0$$