Solve 2x^2+5x+29=0

$$2 { x }^{ 2 } +5x+29=0$$

Solve for x (complex solution)

$$x=\frac{-5+3\sqrt{23}i}{4}\approx -1.25+3.596873642i$$

$$x=\frac{-3\sqrt{23}i-5}{4}\approx -1.25-3.596873642i$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+5x+29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 29}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 29}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\times 29}}{2\times 2}$$

Multiply -8 times 29.

$$x=\frac{-5±\sqrt{25-232}}{2\times 2}$$

Add 25 to -232.

$$x=\frac{-5±\sqrt{-207}}{2\times 2}$$

Take the square root of -207.

$$x=\frac{-5±3\sqrt{23}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±3\sqrt{23}i}{4}$$

Now solve the equation x=\frac{-5±3\sqrt{23}i}{4} when ± is plus. Add -5 to 3i\sqrt{23}.

$$x=\frac{-5+3\sqrt{23}i}{4}$$

Now solve the equation x=\frac{-5±3\sqrt{23}i}{4} when ± is minus. Subtract 3i\sqrt{23} from -5.

$$x=\frac{-3\sqrt{23}i-5}{4}$$

The equation is now solved.

$$x=\frac{-5+3\sqrt{23}i}{4} x=\frac{-3\sqrt{23}i-5}{4}$$

Steps for Completing the Square

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.

$$2x^{2}+5x+29=0$$

Subtract 29 from both sides of the equation.

$$2x^{2}+5x+29-29=-29$$

Subtracting 29 from itself leaves 0.

$$2x^{2}+5x=-29$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=-\frac{29}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=-\frac{29}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-\frac{29}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{29}{2}+\frac{25}{16}$$

Add -\frac{29}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{207}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{207}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{207}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{3\sqrt{23}i}{4} x+\frac{5}{4}=-\frac{3\sqrt{23}i}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{-5+3\sqrt{23}i}{4} x=\frac{-3\sqrt{23}i-5}{4}$$