Solve 4k^2+6k=8k

$$4 k ^ { 2 } + 6 k = 8 k$$

Solve for k

$$k=\frac{1}{2}=0.5$$

$$k=0$$

Steps Using Factoring

Subtract 8k from both sides.

$$4k^{2}+6k-8k=0$$

Combine 6k and -8k to get -2k.

$$4k^{2}-2k=0$$

Factor out k.

$$k\left(4k-2\right)=0$$

To find equation solutions, solve k=0 and 4k-2=0.

$$k=0 k=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 8k from both sides.

$$4k^{2}+6k-8k=0$$

Combine 6k and -8k to get -2k.

$$4k^{2}-2k=0$$

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

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$k=\frac{2±2}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{2±2}{8}$$

Now solve the equation k=\frac{2±2}{8} when ± is plus. Add 2 to 2.

$$k=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$k=\frac{1}{2}$$

Now solve the equation k=\frac{2±2}{8} when ± is minus. Subtract 2 from 2.

$$k=\frac{0}{8}$$

Divide 0 by 8.

$$k=0$$

The equation is now solved.

$$k=\frac{1}{2} k=0$$

Steps for Completing the Square

Subtract 8k from both sides.

$$4k^{2}+6k-8k=0$$

Combine 6k and -8k to get -2k.

$$4k^{2}-2k=0$$

Divide both sides by 4.

$$\frac{4k^{2}-2k}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$k^{2}+\left(-\frac{2}{4}\right)k=\frac{0}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$k^{2}-\frac{1}{2}k=\frac{0}{4}$$

Divide 0 by 4.

$$k^{2}-\frac{1}{2}k=0$$

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.

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

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

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

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}.

$$\left(k-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$k-\frac{1}{4}=\frac{1}{4} k-\frac{1}{4}=-\frac{1}{4}$$

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

$$k=\frac{1}{2} k=0$$