Solve 7x(10-4x)=0

$$7 x ( 10 - 4 x ) = 0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 7x by 10-4x.

$$70x-28x^{2}=0$$

Factor out x.

$$x\left(70-28x\right)=0$$

To find equation solutions, solve x=0 and 70-28x=0.

$$x=0 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7x by 10-4x.

$$70x-28x^{2}=0$$

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.

$$-28x^{2}+70x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -28 for a, 70 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-70±\sqrt{70^{2}}}{2\left(-28\right)}$$

Take the square root of 70^{2}.

$$x=\frac{-70±70}{2\left(-28\right)}$$

Multiply 2 times -28.

$$x=\frac{-70±70}{-56}$$

Now solve the equation x=\frac{-70±70}{-56} when ± is plus. Add -70 to 70.

$$x=\frac{0}{-56}$$

Divide 0 by -56.

$$x=0$$

Now solve the equation x=\frac{-70±70}{-56} when ± is minus. Subtract 70 from -70.

$$x=-\frac{140}{-56}$$

Reduce the fraction \frac{-140}{-56} to lowest terms by extracting and canceling out 28.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=0 x=\frac{5}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 7x by 10-4x.

$$70x-28x^{2}=0$$

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.

$$-28x^{2}+70x=0$$

Divide both sides by -28.

$$\frac{-28x^{2}+70x}{-28}=\frac{0}{-28}$$

Dividing by -28 undoes the multiplication by -28.

$$x^{2}+\frac{70}{-28}x=\frac{0}{-28}$$

Reduce the fraction \frac{70}{-28} to lowest terms by extracting and canceling out 14.

$$x^{2}-\frac{5}{2}x=\frac{0}{-28}$$

Divide 0 by -28.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5}{2} x=0$$