Solve k*2left(2k-2right)=8

$$k \times 2 \left( 2k-2 \right) = 8$$

Solve for k

$$k=-1$$

$$k=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply k\times 2 by 2k-2.

$$4k^{2}-2k\times 2=8$$

Multiply -2 and 2 to get -4.

$$4k^{2}-4k=8$$

Subtract 8 from both sides.

$$4k^{2}-4k-8=0$$

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

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

Square -4.

$$k=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$k=\frac{-\left(-4\right)±\sqrt{16-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$k=\frac{-\left(-4\right)±\sqrt{16+128}}{2\times 4}$$

Add 16 to 128.

$$k=\frac{-\left(-4\right)±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$k=\frac{-\left(-4\right)±12}{2\times 4}$$

The opposite of -4 is 4.

$$k=\frac{4±12}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{4±12}{8}$$

Now solve the equation k=\frac{4±12}{8} when ± is plus. Add 4 to 12.

$$k=\frac{16}{8}$$

Divide 16 by 8.

$$k=2$$

Now solve the equation k=\frac{4±12}{8} when ± is minus. Subtract 12 from 4.

$$k=-\frac{8}{8}$$

Divide -8 by 8.

$$k=-1$$

The equation is now solved.

$$k=2 k=-1$$

Steps for Completing the Square

Use the distributive property to multiply k\times 2 by 2k-2.

$$4k^{2}-2k\times 2=8$$

Multiply -2 and 2 to get -4.

$$4k^{2}-4k=8$$

Divide both sides by 4.

$$\frac{4k^{2}-4k}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$k^{2}+\left(-\frac{4}{4}\right)k=\frac{8}{4}$$

Divide -4 by 4.

$$k^{2}-k=\frac{8}{4}$$

Divide 8 by 4.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-k+\left(-\frac{1}{2}\right)^{2}=2+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-k+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$k^{2}-k+\frac{1}{4}=\frac{9}{4}$$

Factor k^{2}-k+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$k-\frac{1}{2}=\frac{3}{2} k-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$k=2 k=-1$$