Solve 3k(2k-1)=2

$$3 k ( 2 k - 1 ) = 2$$

Solve for k

$$k=\frac{\sqrt{57}}{12}+\frac{1}{4}\approx 0.87915287$$

$$k=-\frac{\sqrt{57}}{12}+\frac{1}{4}\approx -0.37915287$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3k by 2k-1.

$$6k^{2}-3k=2$$

Subtract 2 from both sides.

$$6k^{2}-3k-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Square -3.

$$k=\frac{-\left(-3\right)±\sqrt{9-4\times 6\left(-2\right)}}{2\times 6}$$

Multiply -4 times 6.

$$k=\frac{-\left(-3\right)±\sqrt{9-24\left(-2\right)}}{2\times 6}$$

Multiply -24 times -2.

$$k=\frac{-\left(-3\right)±\sqrt{9+48}}{2\times 6}$$

Add 9 to 48.

$$k=\frac{-\left(-3\right)±\sqrt{57}}{2\times 6}$$

The opposite of -3 is 3.

$$k=\frac{3±\sqrt{57}}{2\times 6}$$

Multiply 2 times 6.

$$k=\frac{3±\sqrt{57}}{12}$$

Now solve the equation k=\frac{3±\sqrt{57}}{12} when ± is plus. Add 3 to \sqrt{57}.

$$k=\frac{\sqrt{57}+3}{12}$$

Divide 3+\sqrt{57} by 12.

$$k=\frac{\sqrt{57}}{12}+\frac{1}{4}$$

Now solve the equation k=\frac{3±\sqrt{57}}{12} when ± is minus. Subtract \sqrt{57} from 3.

$$k=\frac{3-\sqrt{57}}{12}$$

Divide 3-\sqrt{57} by 12.

$$k=-\frac{\sqrt{57}}{12}+\frac{1}{4}$$

The equation is now solved.

$$k=\frac{\sqrt{57}}{12}+\frac{1}{4} k=-\frac{\sqrt{57}}{12}+\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 3k by 2k-1.

$$6k^{2}-3k=2$$

Divide both sides by 6.

$$\frac{6k^{2}-3k}{6}=\frac{2}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$k^{2}+\left(-\frac{3}{6}\right)k=\frac{2}{6}$$

Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.

$$k^{2}-\frac{1}{2}k=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{1}{2}k+\left(-\frac{1}{4}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{1}{2}k+\frac{1}{16}=\frac{1}{3}+\frac{1}{16}$$

Add \frac{1}{3} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{1}{2}k+\frac{1}{16}=\frac{19}{48}$$

Factor k^{2}-\frac{1}{2}k+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{1}{4}\right)^{2}=\frac{19}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{1}{4}\right)^{2}}=\sqrt{\frac{19}{48}}$$

Simplify.

$$k-\frac{1}{4}=\frac{\sqrt{57}}{12} k-\frac{1}{4}=-\frac{\sqrt{57}}{12}$$

Add \frac{1}{4} to both sides of the equation.

$$k=\frac{\sqrt{57}}{12}+\frac{1}{4} k=-\frac{\sqrt{57}}{12}+\frac{1}{4}$$