Solve 5x^2=-9x-6

$$5 x ^ { 2 } = - 9 x - 6$$

Solve for x (complex solution)

$$x=\frac{-9+\sqrt{39}i}{10}\approx -0.9+0.6244998i$$

$$x=\frac{-\sqrt{39}i-9}{10}\approx -0.9-0.6244998i$$

Steps Using the Quadratic Formula

Add 9x to both sides.

$$5x^{2}+9x=-6$$

Add 6 to both sides.

$$5x^{2}+9x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 5\times 6}}{2\times 5}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-9±\sqrt{81-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$x=\frac{-9±\sqrt{81-120}}{2\times 5}$$

Add 81 to -120.

$$x=\frac{-9±\sqrt{-39}}{2\times 5}$$

Take the square root of -39.

$$x=\frac{-9±\sqrt{39}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-9±\sqrt{39}i}{10}$$

Now solve the equation x=\frac{-9±\sqrt{39}i}{10} when ± is plus. Add -9 to i\sqrt{39}.

$$x=\frac{-9+\sqrt{39}i}{10}$$

Now solve the equation x=\frac{-9±\sqrt{39}i}{10} when ± is minus. Subtract i\sqrt{39} from -9.

$$x=\frac{-\sqrt{39}i-9}{10}$$

The equation is now solved.

$$x=\frac{-9+\sqrt{39}i}{10} x=\frac{-\sqrt{39}i-9}{10}$$

Steps for Completing the Square

Add 9x to both sides.

$$5x^{2}+9x=-6$$

Divide both sides by 5.

$$\frac{5x^{2}+9x}{5}=-\frac{6}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{9}{5}x=-\frac{6}{5}$$

Divide \frac{9}{5}, the coefficient of the x term, by 2 to get \frac{9}{10}. Then add the square of \frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{5}x+\left(\frac{9}{10}\right)^{2}=-\frac{6}{5}+\left(\frac{9}{10}\right)^{2}$$

Square \frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{5}x+\frac{81}{100}=-\frac{6}{5}+\frac{81}{100}$$

Add -\frac{6}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{5}x+\frac{81}{100}=-\frac{39}{100}$$

Factor x^{2}+\frac{9}{5}x+\frac{81}{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}.

$$\left(x+\frac{9}{10}\right)^{2}=-\frac{39}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{10}\right)^{2}}=\sqrt{-\frac{39}{100}}$$

Simplify.

$$x+\frac{9}{10}=\frac{\sqrt{39}i}{10} x+\frac{9}{10}=-\frac{\sqrt{39}i}{10}$$

Subtract \frac{9}{10} from both sides of the equation.

$$x=\frac{-9+\sqrt{39}i}{10} x=\frac{-\sqrt{39}i-9}{10}$$