Solve 54x^2+12x+1=0

$$54 { x }^{ 2 } +12x+1=0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{2}i}{18}-\frac{1}{9}\approx -0.111111111+0.07856742i$$

$$x=-\frac{\sqrt{2}i}{18}-\frac{1}{9}\approx -0.111111111-0.07856742i$$

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.

$$54x^{2}+12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, 12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 54}}{2\times 54}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 54}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-12±\sqrt{144-216}}{2\times 54}$$

Add 144 to -216.

$$x=\frac{-12±\sqrt{-72}}{2\times 54}$$

Take the square root of -72.

$$x=\frac{-12±6\sqrt{2}i}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{-12±6\sqrt{2}i}{108}$$

Now solve the equation x=\frac{-12±6\sqrt{2}i}{108} when ± is plus. Add -12 to 6i\sqrt{2}.

$$x=\frac{-12+6\sqrt{2}i}{108}$$

Divide -12+6i\sqrt{2} by 108.

$$x=\frac{\sqrt{2}i}{18}-\frac{1}{9}$$

Now solve the equation x=\frac{-12±6\sqrt{2}i}{108} when ± is minus. Subtract 6i\sqrt{2} from -12.

$$x=\frac{-6\sqrt{2}i-12}{108}$$

Divide -12-6i\sqrt{2} by 108.

$$x=-\frac{\sqrt{2}i}{18}-\frac{1}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{2}i}{18}-\frac{1}{9} x=-\frac{\sqrt{2}i}{18}-\frac{1}{9}$$

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.

$$54x^{2}+12x+1=0$$

Subtract 1 from both sides of the equation.

$$54x^{2}+12x+1-1=-1$$

Subtracting 1 from itself leaves 0.

$$54x^{2}+12x=-1$$

Divide both sides by 54.

$$\frac{54x^{2}+12x}{54}=-\frac{1}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\frac{12}{54}x=-\frac{1}{54}$$

Reduce the fraction \frac{12}{54} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{2}{9}x=-\frac{1}{54}$$

Divide \frac{2}{9}, the coefficient of the x term, by 2 to get \frac{1}{9}. Then add the square of \frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{9}x+\left(\frac{1}{9}\right)^{2}=-\frac{1}{54}+\left(\frac{1}{9}\right)^{2}$$

Square \frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=-\frac{1}{54}+\frac{1}{81}$$

Add -\frac{1}{54} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=-\frac{1}{162}$$

Factor x^{2}+\frac{2}{9}x+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{9}\right)^{2}=-\frac{1}{162}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{9}\right)^{2}}=\sqrt{-\frac{1}{162}}$$

Simplify.

$$x+\frac{1}{9}=\frac{\sqrt{2}i}{18} x+\frac{1}{9}=-\frac{\sqrt{2}i}{18}$$

Subtract \frac{1}{9} from both sides of the equation.

$$x=\frac{\sqrt{2}i}{18}-\frac{1}{9} x=-\frac{\sqrt{2}i}{18}-\frac{1}{9}$$