Solve 216=3k(k+1)

$$216 = 3 k ( k + 1 )$$

Solve for k

$$k=8$$

$$k=-9$$

Steps Using the Quadratic Formula

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

$$216=3k^{2}+3k$$

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

$$3k^{2}+3k=216$$

Subtract 216 from both sides.

$$3k^{2}+3k-216=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 3 for b, and -216 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-3±\sqrt{3^{2}-4\times 3\left(-216\right)}}{2\times 3}$$

Square 3.

$$k=\frac{-3±\sqrt{9-4\times 3\left(-216\right)}}{2\times 3}$$

Multiply -4 times 3.

$$k=\frac{-3±\sqrt{9-12\left(-216\right)}}{2\times 3}$$

Multiply -12 times -216.

$$k=\frac{-3±\sqrt{9+2592}}{2\times 3}$$

Add 9 to 2592.

$$k=\frac{-3±\sqrt{2601}}{2\times 3}$$

Take the square root of 2601.

$$k=\frac{-3±51}{2\times 3}$$

Multiply 2 times 3.

$$k=\frac{-3±51}{6}$$

Now solve the equation k=\frac{-3±51}{6} when ± is plus. Add -3 to 51.

$$k=\frac{48}{6}$$

Divide 48 by 6.

$$k=8$$

Now solve the equation k=\frac{-3±51}{6} when ± is minus. Subtract 51 from -3.

$$k=-\frac{54}{6}$$

Divide -54 by 6.

$$k=-9$$

The equation is now solved.

$$k=8 k=-9$$

Steps for Completing the Square

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

$$216=3k^{2}+3k$$

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

$$3k^{2}+3k=216$$

Divide both sides by 3.

$$\frac{3k^{2}+3k}{3}=\frac{216}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$k^{2}+\frac{3}{3}k=\frac{216}{3}$$

Divide 3 by 3.

$$k^{2}+k=\frac{216}{3}$$

Divide 216 by 3.

$$k^{2}+k=72$$

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.

$$k^{2}+k+\left(\frac{1}{2}\right)^{2}=72+\left(\frac{1}{2}\right)^{2}$$

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

$$k^{2}+k+\frac{1}{4}=72+\frac{1}{4}$$

Add 72 to \frac{1}{4}.

$$k^{2}+k+\frac{1}{4}=\frac{289}{4}$$

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}.

$$\left(k+\frac{1}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{1}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$k+\frac{1}{2}=\frac{17}{2} k+\frac{1}{2}=-\frac{17}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$k=8 k=-9$$