Solve (0-8)(5-12)=(k-12)(4-k)

$$( 0 - 8 ) ( 5 - 12 ) = ( k - 12 ) ( 4 - k )$$

Solve for k

$$k=8+2\sqrt{10}i\approx 8+6.32455532i$$

$$k=-2\sqrt{10}i+8\approx 8-6.32455532i$$

Steps Using the Quadratic Formula

Subtract 8 from 0 to get -8.

$$-8\left(5-12\right)=\left(k-12\right)\left(4-k\right)$$

Subtract 12 from 5 to get -7.

$$-8\left(-7\right)=\left(k-12\right)\left(4-k\right)$$

Multiply -8 and -7 to get 56.

$$56=\left(k-12\right)\left(4-k\right)$$

Use the distributive property to multiply k-12 by 4-k and combine like terms.

$$56=16k-k^{2}-48$$

Swap sides so that all variable terms are on the left hand side.

$$16k-k^{2}-48=56$$

Subtract 56 from both sides.

$$16k-k^{2}-48-56=0$$

Subtract 56 from -48 to get -104.

$$16k-k^{2}-104=0$$

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.

$$-k^{2}+16k-104=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 16 for b, and -104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-16±\sqrt{16^{2}-4\left(-1\right)\left(-104\right)}}{2\left(-1\right)}$$

Square 16.

$$k=\frac{-16±\sqrt{256-4\left(-1\right)\left(-104\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-16±\sqrt{256+4\left(-104\right)}}{2\left(-1\right)}$$

Multiply 4 times -104.

$$k=\frac{-16±\sqrt{256-416}}{2\left(-1\right)}$$

Add 256 to -416.

$$k=\frac{-16±\sqrt{-160}}{2\left(-1\right)}$$

Take the square root of -160.

$$k=\frac{-16±4\sqrt{10}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-16±4\sqrt{10}i}{-2}$$

Now solve the equation k=\frac{-16±4\sqrt{10}i}{-2} when ± is plus. Add -16 to 4i\sqrt{10}.

$$k=\frac{-16+4\sqrt{10}i}{-2}$$

Divide -16+4i\sqrt{10} by -2.

$$k=-2\sqrt{10}i+8$$

Now solve the equation k=\frac{-16±4\sqrt{10}i}{-2} when ± is minus. Subtract 4i\sqrt{10} from -16.

$$k=\frac{-4\sqrt{10}i-16}{-2}$$

Divide -16-4i\sqrt{10} by -2.

$$k=8+2\sqrt{10}i$$

The equation is now solved.

$$k=-2\sqrt{10}i+8 k=8+2\sqrt{10}i$$

Steps for Completing the Square

Subtract 8 from 0 to get -8.

$$-8\left(5-12\right)=\left(k-12\right)\left(4-k\right)$$

Subtract 12 from 5 to get -7.

$$-8\left(-7\right)=\left(k-12\right)\left(4-k\right)$$

Multiply -8 and -7 to get 56.

$$56=\left(k-12\right)\left(4-k\right)$$

Use the distributive property to multiply k-12 by 4-k and combine like terms.

$$56=16k-k^{2}-48$$

Swap sides so that all variable terms are on the left hand side.

$$16k-k^{2}-48=56$$

Add 48 to both sides.

$$16k-k^{2}=56+48$$

Add 56 and 48 to get 104.

$$16k-k^{2}=104$$

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.

$$-k^{2}+16k=104$$

Divide both sides by -1.

$$\frac{-k^{2}+16k}{-1}=\frac{104}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k^{2}+\frac{16}{-1}k=\frac{104}{-1}$$

Divide 16 by -1.

$$k^{2}-16k=\frac{104}{-1}$$

Divide 104 by -1.

$$k^{2}-16k=-104$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-16k+\left(-8\right)^{2}=-104+\left(-8\right)^{2}$$

Square -8.

$$k^{2}-16k+64=-104+64$$

Add -104 to 64.

$$k^{2}-16k+64=-40$$

Factor k^{2}-16k+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-8\right)^{2}=-40$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-8\right)^{2}}=\sqrt{-40}$$

Simplify.

$$k-8=2\sqrt{10}i k-8=-2\sqrt{10}i$$

Add 8 to both sides of the equation.

$$k=8+2\sqrt{10}i k=-2\sqrt{10}i+8$$