Solve 2x-8x(4x+5x)=-38x

$$2x-8x(4x+5x)=-38x$$

Solve for x

$$x=\frac{5}{9}\approx 0.555555556$$

$$x=0$$

Steps Using Factoring

Combine 4x and 5x to get 9x.

$$2x-8x\times 9x=-38x$$

Multiply 8 and 9 to get 72.

$$2x-72xx=-38x$$

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

$$2x-72x^{2}=-38x$$

Add 38x to both sides.

$$2x-72x^{2}+38x=0$$

Combine 2x and 38x to get 40x.

$$40x-72x^{2}=0$$

Factor out x.

$$x\left(40-72x\right)=0$$

To find equation solutions, solve x=0 and 40-72x=0.

$$x=0 x=\frac{5}{9}$$

Steps Using the Quadratic Formula

Combine 4x and 5x to get 9x.

$$2x-8x\times 9x=-38x$$

Multiply 8 and 9 to get 72.

$$2x-72xx=-38x$$

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

$$2x-72x^{2}=-38x$$

Add 38x to both sides.

$$2x-72x^{2}+38x=0$$

Combine 2x and 38x to get 40x.

$$40x-72x^{2}=0$$

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.

$$-72x^{2}+40x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -72 for a, 40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}}}{2\left(-72\right)}$$

Take the square root of 40^{2}.

$$x=\frac{-40±40}{2\left(-72\right)}$$

Multiply 2 times -72.

$$x=\frac{-40±40}{-144}$$

Now solve the equation x=\frac{-40±40}{-144} when ± is plus. Add -40 to 40.

$$x=\frac{0}{-144}$$

Divide 0 by -144.

$$x=0$$

Now solve the equation x=\frac{-40±40}{-144} when ± is minus. Subtract 40 from -40.

$$x=-\frac{80}{-144}$$

Reduce the fraction \frac{-80}{-144} to lowest terms by extracting and canceling out 16.

$$x=\frac{5}{9}$$

The equation is now solved.

$$x=0 x=\frac{5}{9}$$

Steps for Completing the Square

Combine 4x and 5x to get 9x.

$$2x-8x\times 9x=-38x$$

Multiply 8 and 9 to get 72.

$$2x-72xx=-38x$$

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

$$2x-72x^{2}=-38x$$

Add 38x to both sides.

$$2x-72x^{2}+38x=0$$

Combine 2x and 38x to get 40x.

$$40x-72x^{2}=0$$

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.

$$-72x^{2}+40x=0$$

Divide both sides by -72.

$$\frac{-72x^{2}+40x}{-72}=\frac{0}{-72}$$

Dividing by -72 undoes the multiplication by -72.

$$x^{2}+\frac{40}{-72}x=\frac{0}{-72}$$

Reduce the fraction \frac{40}{-72} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{5}{9}x=\frac{0}{-72}$$

Divide 0 by -72.

$$x^{2}-\frac{5}{9}x=0$$

Divide -\frac{5}{9}, the coefficient of the x term, by 2 to get -\frac{5}{18}. Then add the square of -\frac{5}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{9}x+\left(-\frac{5}{18}\right)^{2}=\left(-\frac{5}{18}\right)^{2}$$

Square -\frac{5}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{9}x+\frac{25}{324}=\frac{25}{324}$$

Factor x^{2}-\frac{5}{9}x+\frac{25}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{18}\right)^{2}=\frac{25}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{18}\right)^{2}}=\sqrt{\frac{25}{324}}$$

Simplify.

$$x-\frac{5}{18}=\frac{5}{18} x-\frac{5}{18}=-\frac{5}{18}$$

Add \frac{5}{18} to both sides of the equation.

$$x=\frac{5}{9} x=0$$