Solve left(2x^2right)^+11/20x+1/10=0

$${ \left(2 { x }^{ 2 } \right) }^{ } + \frac{ 11 }{ 20 } x+ \frac{ 1 }{ 10 } = 0$$

Solve for x (complex solution)

$$x=\frac{-11+\sqrt{199}i}{80}\approx -0.1375+0.1763342i$$

$$x=\frac{-\sqrt{199}i-11}{80}\approx -0.1375-0.1763342i$$

Steps Using the Quadratic Formula

Calculate 2x^{2} to the power of 1 and get 2x^{2}.

$$2x^{2}+\frac{11}{20}x+\frac{1}{10}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, \frac{11}{20} for b, and \frac{1}{10} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{20}±\sqrt{\left(\frac{11}{20}\right)^{2}-4\times 2\times \frac{1}{10}}}{2\times 2}$$

Square \frac{11}{20} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{11}{20}±\sqrt{\frac{121}{400}-4\times 2\times \frac{1}{10}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\frac{11}{20}±\sqrt{\frac{121}{400}-8\times \frac{1}{10}}}{2\times 2}$$

Multiply -8 times \frac{1}{10}.

$$x=\frac{-\frac{11}{20}±\sqrt{\frac{121}{400}-\frac{4}{5}}}{2\times 2}$$

Add \frac{121}{400} to -\frac{4}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{11}{20}±\sqrt{-\frac{199}{400}}}{2\times 2}$$

Take the square root of -\frac{199}{400}.

$$x=\frac{-\frac{11}{20}±\frac{\sqrt{199}i}{20}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-\frac{11}{20}±\frac{\sqrt{199}i}{20}}{4}$$

Now solve the equation x=\frac{-\frac{11}{20}±\frac{\sqrt{199}i}{20}}{4} when ± is plus. Add -\frac{11}{20} to \frac{i\sqrt{199}}{20}.

$$x=\frac{-11+\sqrt{199}i}{4\times 20}$$

Divide \frac{-11+i\sqrt{199}}{20} by 4.

$$x=\frac{-11+\sqrt{199}i}{80}$$

Now solve the equation x=\frac{-\frac{11}{20}±\frac{\sqrt{199}i}{20}}{4} when ± is minus. Subtract \frac{i\sqrt{199}}{20} from -\frac{11}{20}.

$$x=\frac{-\sqrt{199}i-11}{4\times 20}$$

Divide \frac{-11-i\sqrt{199}}{20} by 4.

$$x=\frac{-\sqrt{199}i-11}{80}$$

The equation is now solved.

$$x=\frac{-11+\sqrt{199}i}{80} x=\frac{-\sqrt{199}i-11}{80}$$

Steps for Completing the Square

Calculate 2x^{2} to the power of 1 and get 2x^{2}.

$$2x^{2}+\frac{11}{20}x+\frac{1}{10}=0$$

Subtract \frac{1}{10} from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+\frac{11}{20}x=-\frac{1}{10}$$

Divide both sides by 2.

$$\frac{2x^{2}+\frac{11}{20}x}{2}=-\frac{\frac{1}{10}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{\frac{11}{20}}{2}x=-\frac{\frac{1}{10}}{2}$$

Divide \frac{11}{20} by 2.

$$x^{2}+\frac{11}{40}x=-\frac{\frac{1}{10}}{2}$$

Divide -\frac{1}{10} by 2.

$$x^{2}+\frac{11}{40}x=-\frac{1}{20}$$

Divide \frac{11}{40}, the coefficient of the x term, by 2 to get \frac{11}{80}. Then add the square of \frac{11}{80} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{40}x+\left(\frac{11}{80}\right)^{2}=-\frac{1}{20}+\left(\frac{11}{80}\right)^{2}$$

Square \frac{11}{80} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{40}x+\frac{121}{6400}=-\frac{1}{20}+\frac{121}{6400}$$

Add -\frac{1}{20} to \frac{121}{6400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{40}x+\frac{121}{6400}=-\frac{199}{6400}$$

Factor x^{2}+\frac{11}{40}x+\frac{121}{6400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{80}\right)^{2}=-\frac{199}{6400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{80}\right)^{2}}=\sqrt{-\frac{199}{6400}}$$

Simplify.

$$x+\frac{11}{80}=\frac{\sqrt{199}i}{80} x+\frac{11}{80}=-\frac{\sqrt{199}i}{80}$$

Subtract \frac{11}{80} from both sides of the equation.

$$x=\frac{-11+\sqrt{199}i}{80} x=\frac{-\sqrt{199}i-11}{80}$$