Solve -4k(5k-6)=0

$$- 4 k ( 5 k - 6 ) = 0$$

Solve for k

$$k = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

$$k=0$$

Steps Using Factoring

Use the distributive property to multiply -4k by 5k-6.

$$-20k^{2}+24k=0$$

Factor out k.

$$k\left(-20k+24\right)=0$$

To find equation solutions, solve k=0 and -20k+24=0.

$$k=0 k=\frac{6}{5}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4k by 5k-6.

$$-20k^{2}+24k=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 24 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-24±\sqrt{24^{2}}}{2\left(-20\right)}$$

Take the square root of 24^{2}.

$$k=\frac{-24±24}{2\left(-20\right)}$$

Multiply 2 times -20.

$$k=\frac{-24±24}{-40}$$

Now solve the equation k=\frac{-24±24}{-40} when ± is plus. Add -24 to 24.

$$k=\frac{0}{-40}$$

Divide 0 by -40.

$$k=0$$

Now solve the equation k=\frac{-24±24}{-40} when ± is minus. Subtract 24 from -24.

$$k=-\frac{48}{-40}$$

Reduce the fraction \frac{-48}{-40} to lowest terms by extracting and canceling out 8.

$$k=\frac{6}{5}$$

The equation is now solved.

$$k=0 k=\frac{6}{5}$$

Steps for Completing the Square

Use the distributive property to multiply -4k by 5k-6.

$$-20k^{2}+24k=0$$

Divide both sides by -20.

$$\frac{-20k^{2}+24k}{-20}=\frac{0}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$k^{2}+\frac{24}{-20}k=\frac{0}{-20}$$

Reduce the fraction \frac{24}{-20} to lowest terms by extracting and canceling out 4.

$$k^{2}-\frac{6}{5}k=\frac{0}{-20}$$

Divide 0 by -20.

$$k^{2}-\frac{6}{5}k=0$$

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

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

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{6}{5}k+\frac{9}{25}=\frac{9}{25}$$

Factor k^{2}-\frac{6}{5}k+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{3}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{3}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

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

Add \frac{3}{5} to both sides of the equation.

$$k=\frac{6}{5} k=0$$