Solve 34k^2+18k=0

$$34 k ^ { 2 } + 18 k = 0$$

Solve for k

$$k=-\frac{9}{17}\approx -0.529411765$$

$$k=0$$

Steps Using Factoring

Factor out k.

$$k\left(34k+18\right)=0$$

To find equation solutions, solve k=0 and 34k+18=0.

$$k=0 k=-\frac{9}{17}$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$34k^{2}+18k=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 34 for a, 18 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-18±\sqrt{18^{2}}}{2\times 34}$$

Take the square root of 18^{2}.

$$k=\frac{-18±18}{2\times 34}$$

Multiply 2 times 34.

$$k=\frac{-18±18}{68}$$

Now solve the equation k=\frac{-18±18}{68} when ± is plus. Add -18 to 18.

$$k=\frac{0}{68}$$

Divide 0 by 68.

$$k=0$$

Now solve the equation k=\frac{-18±18}{68} when ± is minus. Subtract 18 from -18.

$$k=-\frac{36}{68}$$

Reduce the fraction \frac{-36}{68} to lowest terms by extracting and canceling out 4.

$$k=-\frac{9}{17}$$

The equation is now solved.

$$k=0 k=-\frac{9}{17}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$34k^{2}+18k=0$$

Divide both sides by 34.

$$\frac{34k^{2}+18k}{34}=\frac{0}{34}$$

Dividing by 34 undoes the multiplication by 34.

$$k^{2}+\frac{18}{34}k=\frac{0}{34}$$

Reduce the fraction \frac{18}{34} to lowest terms by extracting and canceling out 2.

$$k^{2}+\frac{9}{17}k=\frac{0}{34}$$

Divide 0 by 34.

$$k^{2}+\frac{9}{17}k=0$$

Divide \frac{9}{17}, the coefficient of the x term, by 2 to get \frac{9}{34}. Then add the square of \frac{9}{34} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{9}{17}k+\left(\frac{9}{34}\right)^{2}=\left(\frac{9}{34}\right)^{2}$$

Square \frac{9}{34} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{9}{17}k+\frac{81}{1156}=\frac{81}{1156}$$

Factor k^{2}+\frac{9}{17}k+\frac{81}{1156}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{9}{34}\right)^{2}=\frac{81}{1156}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{9}{34}\right)^{2}}=\sqrt{\frac{81}{1156}}$$

Simplify.

$$k+\frac{9}{34}=\frac{9}{34} k+\frac{9}{34}=-\frac{9}{34}$$

Subtract \frac{9}{34} from both sides of the equation.

$$k=0 k=-\frac{9}{17}$$