Solve x*(7x-5)=18

$$x \cdot ( 7 x - 5 ) = 18$$

Solve for x

$$x = -\frac{9}{7} = -1\frac{2}{7} \approx -1.285714286$$

$$x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 7x-5.

$$7x^{2}-5x=18$$

Subtract 18 from both sides.

$$7x^{2}-5x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -5 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 7\left(-18\right)}}{2\times 7}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 7\left(-18\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-5\right)±\sqrt{25-28\left(-18\right)}}{2\times 7}$$

Multiply -28 times -18.

$$x=\frac{-\left(-5\right)±\sqrt{25+504}}{2\times 7}$$

Add 25 to 504.

$$x=\frac{-\left(-5\right)±\sqrt{529}}{2\times 7}$$

Take the square root of 529.

$$x=\frac{-\left(-5\right)±23}{2\times 7}$$

The opposite of -5 is 5.

$$x=\frac{5±23}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{5±23}{14}$$

Now solve the equation x=\frac{5±23}{14} when ± is plus. Add 5 to 23.

$$x=\frac{28}{14}$$

Divide 28 by 14.

$$x=2$$

Now solve the equation x=\frac{5±23}{14} when ± is minus. Subtract 23 from 5.

$$x=-\frac{18}{14}$$

Reduce the fraction \frac{-18}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{7}$$

The equation is now solved.

$$x=2 x=-\frac{9}{7}$$

Steps for Completing the Square

Use the distributive property to multiply x by 7x-5.

$$7x^{2}-5x=18$$

Divide both sides by 7.

$$\frac{7x^{2}-5x}{7}=\frac{18}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{5}{7}x=\frac{18}{7}$$

Divide -\frac{5}{7}, the coefficient of the x term, by 2 to get -\frac{5}{14}. Then add the square of -\frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{7}x+\left(-\frac{5}{14}\right)^{2}=\frac{18}{7}+\left(-\frac{5}{14}\right)^{2}$$

Square -\frac{5}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{7}x+\frac{25}{196}=\frac{18}{7}+\frac{25}{196}$$

Add \frac{18}{7} to \frac{25}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{7}x+\frac{25}{196}=\frac{529}{196}$$

Factor x^{2}-\frac{5}{7}x+\frac{25}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{14}\right)^{2}=\frac{529}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{14}\right)^{2}}=\sqrt{\frac{529}{196}}$$

Simplify.

$$x-\frac{5}{14}=\frac{23}{14} x-\frac{5}{14}=-\frac{23}{14}$$

Add \frac{5}{14} to both sides of the equation.

$$x=2 x=-\frac{9}{7}$$