Solve left(2xright)^2+5x+6=0

$${ \left(2x \right) }^{ 2 } +5x+6=0$$

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{71}i}{8}\approx -0.625+1.053268722i$$

$$x=\frac{-\sqrt{71}i-5}{8}\approx -0.625-1.053268722i$$

Steps Using the Quadratic Formula

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+5x+6=0$$

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-5±\sqrt{25-96}}{2\times 4}$$

Add 25 to -96.

$$x=\frac{-5±\sqrt{-71}}{2\times 4}$$

Take the square root of -71.

$$x=\frac{-5±\sqrt{71}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±\sqrt{71}i}{8}$$

Now solve the equation x=\frac{-5±\sqrt{71}i}{8} when ± is plus. Add -5 to i\sqrt{71}.

$$x=\frac{-5+\sqrt{71}i}{8}$$

Now solve the equation x=\frac{-5±\sqrt{71}i}{8} when ± is minus. Subtract i\sqrt{71} from -5.

$$x=\frac{-\sqrt{71}i-5}{8}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{71}i}{8} x=\frac{-\sqrt{71}i-5}{8}$$

Steps for Completing the Square

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+5x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+5x=-6$$

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=-\frac{6}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{4}x=-\frac{3}{2}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=-\frac{3}{2}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{3}{2}+\frac{25}{64}$$

Add -\frac{3}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{71}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=-\frac{71}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{-\frac{71}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{71}i}{8} x+\frac{5}{8}=-\frac{\sqrt{71}i}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{-5+\sqrt{71}i}{8} x=\frac{-\sqrt{71}i-5}{8}$$