Solve (4-k)(k-5)=-42

$$( 4 - k ) ( k - 5 ) = - 42$$

Solve for k

$$k=-2$$

$$k=11$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4-k by k-5 and combine like terms.

$$9k-20-k^{2}=-42$$

Add 42 to both sides.

$$9k-20-k^{2}+42=0$$

Add -20 and 42 to get 22.

$$9k+22-k^{2}=0$$

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.

$$-k^{2}+9k+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\times 22}}{2\left(-1\right)}$$

Square 9.

$$k=\frac{-9±\sqrt{81-4\left(-1\right)\times 22}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-9±\sqrt{81+4\times 22}}{2\left(-1\right)}$$

Multiply 4 times 22.

$$k=\frac{-9±\sqrt{81+88}}{2\left(-1\right)}$$

Add 81 to 88.

$$k=\frac{-9±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$k=\frac{-9±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-9±13}{-2}$$

Now solve the equation k=\frac{-9±13}{-2} when ± is plus. Add -9 to 13.

$$k=\frac{4}{-2}$$

Divide 4 by -2.

$$k=-2$$

Now solve the equation k=\frac{-9±13}{-2} when ± is minus. Subtract 13 from -9.

$$k=-\frac{22}{-2}$$

Divide -22 by -2.

$$k=11$$

The equation is now solved.

$$k=-2 k=11$$

Steps for Completing the Square

Use the distributive property to multiply 4-k by k-5 and combine like terms.

$$9k-20-k^{2}=-42$$

Add 20 to both sides.

$$9k-k^{2}=-42+20$$

Add -42 and 20 to get -22.

$$9k-k^{2}=-22$$

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.

$$-k^{2}+9k=-22$$

Divide both sides by -1.

$$\frac{-k^{2}+9k}{-1}=-\frac{22}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k^{2}+\frac{9}{-1}k=-\frac{22}{-1}$$

Divide 9 by -1.

$$k^{2}-9k=-\frac{22}{-1}$$

Divide -22 by -1.

$$k^{2}-9k=22$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-9k+\left(-\frac{9}{2}\right)^{2}=22+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-9k+\frac{81}{4}=22+\frac{81}{4}$$

Add 22 to \frac{81}{4}.

$$k^{2}-9k+\frac{81}{4}=\frac{169}{4}$$

Factor k^{2}-9k+\frac{81}{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}.

$$\left(k-\frac{9}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{9}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$k-\frac{9}{2}=\frac{13}{2} k-\frac{9}{2}=-\frac{13}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$k=11 k=-2$$