Solve 2x^2=-7x-29

$$2 x ^ { 2 } = - 7 x - 29$$

Solve for x (complex solution)

$$x=\frac{-7+\sqrt{183}i}{4}\approx -1.75+3.381937315i$$

$$x=\frac{-\sqrt{183}i-7}{4}\approx -1.75-3.381937315i$$

Steps Using the Quadratic Formula

Add 7x to both sides.

$$2x^{2}+7x=-29$$

Add 29 to both sides.

$$2x^{2}+7x+29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 29}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 29}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 29}}{2\times 2}$$

Multiply -8 times 29.

$$x=\frac{-7±\sqrt{49-232}}{2\times 2}$$

Add 49 to -232.

$$x=\frac{-7±\sqrt{-183}}{2\times 2}$$

Take the square root of -183.

$$x=\frac{-7±\sqrt{183}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±\sqrt{183}i}{4}$$

Now solve the equation x=\frac{-7±\sqrt{183}i}{4} when ± is plus. Add -7 to i\sqrt{183}.

$$x=\frac{-7+\sqrt{183}i}{4}$$

Now solve the equation x=\frac{-7±\sqrt{183}i}{4} when ± is minus. Subtract i\sqrt{183} from -7.

$$x=\frac{-\sqrt{183}i-7}{4}$$

The equation is now solved.

$$x=\frac{-7+\sqrt{183}i}{4} x=\frac{-\sqrt{183}i-7}{4}$$

Steps for Completing the Square

Add 7x to both sides.

$$2x^{2}+7x=-29$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=-\frac{29}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=-\frac{29}{2}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-\frac{29}{2}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{29}{2}+\frac{49}{16}$$

Add -\frac{29}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{183}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{4}\right)^{2}=-\frac{183}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{-\frac{183}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{\sqrt{183}i}{4} x+\frac{7}{4}=-\frac{\sqrt{183}i}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{-7+\sqrt{183}i}{4} x=\frac{-\sqrt{183}i-7}{4}$$