Solve 6x^2-10x=8x

$$6 { x }^{ 2 } -10x=8x$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Subtract 8x from both sides.

$$6x^{2}-10x-8x=0$$

Combine -10x and -8x to get -18x.

$$6x^{2}-18x=0$$

Factor out x.

$$x\left(6x-18\right)=0$$

To find equation solutions, solve x=0 and 6x-18=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$6x^{2}-10x-8x=0$$

Combine -10x and -8x to get -18x.

$$6x^{2}-18x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -18 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}}}{2\times 6}$$

Take the square root of \left(-18\right)^{2}.

$$x=\frac{-\left(-18\right)±18}{2\times 6}$$

The opposite of -18 is 18.

$$x=\frac{18±18}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{18±18}{12}$$

Now solve the equation x=\frac{18±18}{12} when ± is plus. Add 18 to 18.

$$x=\frac{36}{12}$$

Divide 36 by 12.

$$x=3$$

Now solve the equation x=\frac{18±18}{12} when ± is minus. Subtract 18 from 18.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

The equation is now solved.

$$x=3 x=0$$

Steps for Completing the Square

Subtract 8x from both sides.

$$6x^{2}-10x-8x=0$$

Combine -10x and -8x to get -18x.

$$6x^{2}-18x=0$$

Divide both sides by 6.

$$\frac{6x^{2}-18x}{6}=\frac{0}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{18}{6}\right)x=\frac{0}{6}$$

Divide -18 by 6.

$$x^{2}-3x=\frac{0}{6}$$

Divide 0 by 6.

$$x^{2}-3x=0$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$