Solve 5x^2-2x^-17=0

$$5 { x }^{ 2 } -2 { x }^{ } -17=0$$

Solve for x

$$x = \frac{\sqrt{86} + 1}{5} \approx 2.054723699$$

$$x=\frac{1-\sqrt{86}}{5}\approx -1.654723699$$

Steps Using the Quadratic Formula

Calculate x to the power of 1 and get x.

$$5x^{2}-2x-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -2 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\left(-17\right)}}{2\times 5}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 5\left(-17\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-20\left(-17\right)}}{2\times 5}$$

Multiply -20 times -17.

$$x=\frac{-\left(-2\right)±\sqrt{4+340}}{2\times 5}$$

Add 4 to 340.

$$x=\frac{-\left(-2\right)±\sqrt{344}}{2\times 5}$$

Take the square root of 344.

$$x=\frac{-\left(-2\right)±2\sqrt{86}}{2\times 5}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{86}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{2±2\sqrt{86}}{10}$$

Now solve the equation x=\frac{2±2\sqrt{86}}{10} when ± is plus. Add 2 to 2\sqrt{86}.

$$x=\frac{2\sqrt{86}+2}{10}$$

Divide 2+2\sqrt{86} by 10.

$$x=\frac{\sqrt{86}+1}{5}$$

Now solve the equation x=\frac{2±2\sqrt{86}}{10} when ± is minus. Subtract 2\sqrt{86} from 2.

$$x=\frac{2-2\sqrt{86}}{10}$$

Divide 2-2\sqrt{86} by 10.

$$x=\frac{1-\sqrt{86}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{86}+1}{5} x=\frac{1-\sqrt{86}}{5}$$

Steps for Completing the Square

Calculate x to the power of 1 and get x.

$$5x^{2}-2x-17=0$$

Add 17 to both sides. Anything plus zero gives itself.

$$5x^{2}-2x=17$$

Divide both sides by 5.

$$\frac{5x^{2}-2x}{5}=\frac{17}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{2}{5}x=\frac{17}{5}$$

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

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{17}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{17}{5}+\frac{1}{25}$$

Add \frac{17}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{86}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{86}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{86}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{\sqrt{86}}{5} x-\frac{1}{5}=-\frac{\sqrt{86}}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{\sqrt{86}+1}{5} x=\frac{1-\sqrt{86}}{5}$$