Solve x^2*5=7x

$$x ^ { 2 } \times 5 = 7 x$$

Solve for x

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

$$x=0$$

Steps Using Factoring

Subtract 7x from both sides.

$$x^{2}\times 5-7x=0$$

Factor out x.

$$x\left(5x-7\right)=0$$

To find equation solutions, solve x=0 and 5x-7=0.

$$x=0 x=\frac{7}{5}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$x^{2}\times 5-7x=0$$

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.

$$5x^{2}-7x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}}}{2\times 5}$$

Take the square root of \left(-7\right)^{2}.

$$x=\frac{-\left(-7\right)±7}{2\times 5}$$

The opposite of -7 is 7.

$$x=\frac{7±7}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{7±7}{10}$$

Now solve the equation x=\frac{7±7}{10} when ± is plus. Add 7 to 7.

$$x=\frac{14}{10}$$

Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{5}$$

Now solve the equation x=\frac{7±7}{10} when ± is minus. Subtract 7 from 7.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{7}{5} x=0$$

Steps for Completing the Square

Subtract 7x from both sides.

$$x^{2}\times 5-7x=0$$

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.

$$5x^{2}-7x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-7x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{7}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{7}{5}x=0$$

Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{5}x+\left(-\frac{7}{10}\right)^{2}=\left(-\frac{7}{10}\right)^{2}$$

Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{49}{100}$$

Factor x^{2}-\frac{7}{5}x+\frac{49}{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}.

$$\left(x-\frac{7}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x-\frac{7}{10}=\frac{7}{10} x-\frac{7}{10}=-\frac{7}{10}$$

Add \frac{7}{10} to both sides of the equation.

$$x=\frac{7}{5} x=0$$