Evaluate: 5 x ^ 2-x = 9

Expression: $$5 x ^ { 2 } - x = 9$$

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 $x^{2}+bx=c$.

$$5x^{2}-x=9$$

Divide both sides by $5$.

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

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\frac{-1}{5}x=\frac{9}{5}$$

Divide $-1$ by $5$.

$$x^{2}-\frac{1}{5}x=\frac{9}{5}$$

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.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{9}{5}+\left(-\frac{1}{10}\right)^{2}$$

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

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{9}{5}+\frac{1}{100}$$

Add $\frac{9}{5}$ to $\frac{1}{100}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{181}{100}$$

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 $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{181}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{181}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{\sqrt{181}}{10}$$ $$x-\frac{1}{10}=-\frac{\sqrt{181}}{10}$$

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

$$x=\frac{\sqrt{181}+1}{10}$$ $$x=\frac{1-\sqrt{181}}{10}$$