Solve 5x(3-4x)=15x-4x

$$5 x ( 3 - 4 x ) = 15 x - 4 x$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 5x by 3-4x.

$$15x-20x^{2}=15x-4x$$

Combine 15x and -4x to get 11x.

$$15x-20x^{2}=11x$$

Subtract 11x from both sides.

$$15x-20x^{2}-11x=0$$

Combine 15x and -11x to get 4x.

$$4x-20x^{2}=0$$

Factor out x.

$$x\left(4-20x\right)=0$$

To find equation solutions, solve x=0 and 4-20x=0.

$$x=0 x=\frac{1}{5}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by 3-4x.

$$15x-20x^{2}=15x-4x$$

Combine 15x and -4x to get 11x.

$$15x-20x^{2}=11x$$

Subtract 11x from both sides.

$$15x-20x^{2}-11x=0$$

Combine 15x and -11x to get 4x.

$$4x-20x^{2}=0$$

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.

$$-20x^{2}+4x=0$$

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

$$x=\frac{-4±\sqrt{4^{2}}}{2\left(-20\right)}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-4±4}{-40}$$

Now solve the equation x=\frac{-4±4}{-40} when ± is plus. Add -4 to 4.

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

Divide 0 by -40.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-40} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-40}$$

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

$$x=\frac{1}{5}$$

The equation is now solved.

$$x=0 x=\frac{1}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 5x by 3-4x.

$$15x-20x^{2}=15x-4x$$

Combine 15x and -4x to get 11x.

$$15x-20x^{2}=11x$$

Subtract 11x from both sides.

$$15x-20x^{2}-11x=0$$

Combine 15x and -11x to get 4x.

$$4x-20x^{2}=0$$

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.

$$-20x^{2}+4x=0$$

Divide both sides by -20.

$$\frac{-20x^{2}+4x}{-20}=\frac{0}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$x^{2}+\frac{4}{-20}x=\frac{0}{-20}$$

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

$$x^{2}-\frac{1}{5}x=\frac{0}{-20}$$

Divide 0 by -20.

$$x^{2}-\frac{1}{5}x=0$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{1}{10} x-\frac{1}{10}=-\frac{1}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{1}{5} x=0$$