Solve 0=-9x^2-2x+8

$$0 = - 9 x ^ { 2 } - 2 x + 8$$

Solve for x

$$x=\frac{\sqrt{73}-1}{9}\approx 0.838222638$$

$$x=\frac{-\sqrt{73}-1}{9}\approx -1.060444861$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-9x^{2}-2x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -2 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-9\right)\times 8}}{2\left(-9\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-9\right)\times 8}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-2\right)±\sqrt{4+36\times 8}}{2\left(-9\right)}$$

Multiply 36 times 8.

$$x=\frac{-\left(-2\right)±\sqrt{4+288}}{2\left(-9\right)}$$

Add 4 to 288.

$$x=\frac{-\left(-2\right)±\sqrt{292}}{2\left(-9\right)}$$

Take the square root of 292.

$$x=\frac{-\left(-2\right)±2\sqrt{73}}{2\left(-9\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{73}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{2±2\sqrt{73}}{-18}$$

Now solve the equation x=\frac{2±2\sqrt{73}}{-18} when ± is plus. Add 2 to 2\sqrt{73}.

$$x=\frac{2\sqrt{73}+2}{-18}$$

Divide 2+2\sqrt{73} by -18.

$$x=\frac{-\sqrt{73}-1}{9}$$

Now solve the equation x=\frac{2±2\sqrt{73}}{-18} when ± is minus. Subtract 2\sqrt{73} from 2.

$$x=\frac{2-2\sqrt{73}}{-18}$$

Divide 2-2\sqrt{73} by -18.

$$x=\frac{\sqrt{73}-1}{9}$$

The equation is now solved.

$$x=\frac{-\sqrt{73}-1}{9} x=\frac{\sqrt{73}-1}{9}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-9x^{2}-2x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-9x^{2}-2x=-8$$

Divide both sides by -9.

$$\frac{-9x^{2}-2x}{-9}=-\frac{8}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{2}{-9}\right)x=-\frac{8}{-9}$$

Divide -2 by -9.

$$x^{2}+\frac{2}{9}x=-\frac{8}{-9}$$

Divide -8 by -9.

$$x^{2}+\frac{2}{9}x=\frac{8}{9}$$

Divide \frac{2}{9}, the coefficient of the x term, by 2 to get \frac{1}{9}. Then add the square of \frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{9}x+\left(\frac{1}{9}\right)^{2}=\frac{8}{9}+\left(\frac{1}{9}\right)^{2}$$

Square \frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=\frac{8}{9}+\frac{1}{81}$$

Add \frac{8}{9} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=\frac{73}{81}$$

Factor x^{2}+\frac{2}{9}x+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{9}\right)^{2}=\frac{73}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{9}\right)^{2}}=\sqrt{\frac{73}{81}}$$

Simplify.

$$x+\frac{1}{9}=\frac{\sqrt{73}}{9} x+\frac{1}{9}=-\frac{\sqrt{73}}{9}$$

Subtract \frac{1}{9} from both sides of the equation.

$$x=\frac{\sqrt{73}-1}{9} x=\frac{-\sqrt{73}-1}{9}$$