Solve (3k^2-k)+2(3k+1)/2=11

$$\frac { ( 3 k ^ { 2 } - k ) + 2 ( 3 k + 1 ) } { 2 } = 11$$

Solve for k

$$k = \frac{\sqrt{265} - 5}{6} \approx 1.879803433$$

$$k=\frac{-\sqrt{265}-5}{6}\approx -3.546470099$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$3k^{2}-k+2\left(3k+1\right)=11\times 2$$

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

$$3k^{2}-k+6k+2=11\times 2$$

Combine -k and 6k to get 5k.

$$3k^{2}+5k+2=11\times 2$$

Multiply 11 and 2 to get 22.

$$3k^{2}+5k+2=22$$

Subtract 22 from both sides.

$$3k^{2}+5k+2-22=0$$

Subtract 22 from 2 to get -20.

$$3k^{2}+5k-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-5±\sqrt{5^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Square 5.

$$k=\frac{-5±\sqrt{25-4\times 3\left(-20\right)}}{2\times 3}$$

Multiply -4 times 3.

$$k=\frac{-5±\sqrt{25-12\left(-20\right)}}{2\times 3}$$

Multiply -12 times -20.

$$k=\frac{-5±\sqrt{25+240}}{2\times 3}$$

Add 25 to 240.

$$k=\frac{-5±\sqrt{265}}{2\times 3}$$

Multiply 2 times 3.

$$k=\frac{-5±\sqrt{265}}{6}$$

Now solve the equation k=\frac{-5±\sqrt{265}}{6} when ± is plus. Add -5 to \sqrt{265}.

$$k=\frac{\sqrt{265}-5}{6}$$

Now solve the equation k=\frac{-5±\sqrt{265}}{6} when ± is minus. Subtract \sqrt{265} from -5.

$$k=\frac{-\sqrt{265}-5}{6}$$

The equation is now solved.

$$k=\frac{\sqrt{265}-5}{6} k=\frac{-\sqrt{265}-5}{6}$$

Steps for Completing the Square

Multiply both sides by 2.

$$3k^{2}-k+2\left(3k+1\right)=11\times 2$$

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

$$3k^{2}-k+6k+2=11\times 2$$

Combine -k and 6k to get 5k.

$$3k^{2}+5k+2=11\times 2$$

Multiply 11 and 2 to get 22.

$$3k^{2}+5k+2=22$$

Subtract 2 from both sides.

$$3k^{2}+5k=22-2$$

Subtract 2 from 22 to get 20.

$$3k^{2}+5k=20$$

Divide both sides by 3.

$$\frac{3k^{2}+5k}{3}=\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$k^{2}+\frac{5}{3}k=\frac{20}{3}$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{5}{3}k+\left(\frac{5}{6}\right)^{2}=\frac{20}{3}+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{5}{3}k+\frac{25}{36}=\frac{20}{3}+\frac{25}{36}$$

Add \frac{20}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{5}{3}k+\frac{25}{36}=\frac{265}{36}$$

Factor k^{2}+\frac{5}{3}k+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{5}{6}\right)^{2}=\frac{265}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{5}{6}\right)^{2}}=\sqrt{\frac{265}{36}}$$

Simplify.

$$k+\frac{5}{6}=\frac{\sqrt{265}}{6} k+\frac{5}{6}=-\frac{\sqrt{265}}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$k=\frac{\sqrt{265}-5}{6} k=\frac{-\sqrt{265}-5}{6}$$