Solve 2k^2-5k+2=k

$$2 k ^ { 2 } - 5 k + 2 = k$$

Solve for k

$$k = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

$$k=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

Steps Using the Quadratic Formula

Subtract k from both sides.

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

Combine -5k and -k to get -6k.

$$2k^{2}-6k+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square -6.

$$k=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-\left(-6\right)±\sqrt{36-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$k=\frac{-\left(-6\right)±\sqrt{36-16}}{2\times 2}$$

Add 36 to -16.

$$k=\frac{-\left(-6\right)±\sqrt{20}}{2\times 2}$$

Take the square root of 20.

$$k=\frac{-\left(-6\right)±2\sqrt{5}}{2\times 2}$$

The opposite of -6 is 6.

$$k=\frac{6±2\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{6±2\sqrt{5}}{4}$$

Now solve the equation k=\frac{6±2\sqrt{5}}{4} when ± is plus. Add 6 to 2\sqrt{5}.

$$k=\frac{2\sqrt{5}+6}{4}$$

Divide 6+2\sqrt{5} by 4.

$$k=\frac{\sqrt{5}+3}{2}$$

Now solve the equation k=\frac{6±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from 6.

$$k=\frac{6-2\sqrt{5}}{4}$$

Divide 6-2\sqrt{5} by 4.

$$k=\frac{3-\sqrt{5}}{2}$$

The equation is now solved.

$$k=\frac{\sqrt{5}+3}{2} k=\frac{3-\sqrt{5}}{2}$$

Steps for Completing the Square

Subtract k from both sides.

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

Combine -5k and -k to get -6k.

$$2k^{2}-6k+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2k^{2}-6k=-2$$

Divide both sides by 2.

$$\frac{2k^{2}-6k}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\left(-\frac{6}{2}\right)k=-\frac{2}{2}$$

Divide -6 by 2.

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

Divide -2 by 2.

$$k^{2}-3k=-1$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-3k+\left(-\frac{3}{2}\right)^{2}=-1+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-3k+\frac{9}{4}=-1+\frac{9}{4}$$

Add -1 to \frac{9}{4}.

$$k^{2}-3k+\frac{9}{4}=\frac{5}{4}$$

Factor k^{2}-3k+\frac{9}{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}.

$$\left(k-\frac{3}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$k-\frac{3}{2}=\frac{\sqrt{5}}{2} k-\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$k=\frac{\sqrt{5}+3}{2} k=\frac{3-\sqrt{5}}{2}$$