Solve x^2-4xx+32x=0

$$x ^ { 2 } - 4 x x + 32 x = 0$$

Solve for x

$$x = \frac{32}{3} = 10\frac{2}{3} \approx 10.666666667$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(x-4x+32\right)=0$$

To find equation solutions, solve x=0 and x-4x+32=0.

$$x=0 x=\frac{32}{3}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$x^{2}-4x^{2}+32x=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+32x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 32 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}}}{2\left(-3\right)}$$

Take the square root of 32^{2}.

$$x=\frac{-32±32}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-32±32}{-6}$$

Now solve the equation x=\frac{-32±32}{-6} when ± is plus. Add -32 to 32.

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-32±32}{-6} when ± is minus. Subtract 32 from -32.

$$x=-\frac{64}{-6}$$

Reduce the fraction \frac{-64}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{32}{3}$$

The equation is now solved.

$$x=0 x=\frac{32}{3}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$x^{2}-4x^{2}+32x=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+32x=0$$

Divide both sides by -3.

$$\frac{-3x^{2}+32x}{-3}=\frac{0}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{32}{-3}x=\frac{0}{-3}$$

Divide 32 by -3.

$$x^{2}-\frac{32}{3}x=\frac{0}{-3}$$

Divide 0 by -3.

$$x^{2}-\frac{32}{3}x=0$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{256}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{256}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{256}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{16}{3} x-\frac{16}{3}=-\frac{16}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=\frac{32}{3} x=0$$