Solve 5x^2=4x-1

$$5 x ^ { 2 } = 4 x - 1$$

Solve for x (complex solution)

$$x=\frac{2}{5}+\frac{1}{5}i=0.4+0.2i$$

$$x=\frac{2}{5}-\frac{1}{5}i=0.4-0.2i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$5x^{2}-4x=-1$$

Add 1 to both sides.

$$5x^{2}-4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5}}{2\times 5}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20}}{2\times 5}$$

Add 16 to -20.

$$x=\frac{-\left(-4\right)±\sqrt{-4}}{2\times 5}$$

Take the square root of -4.

$$x=\frac{-\left(-4\right)±2i}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±2i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±2i}{10}$$

Now solve the equation x=\frac{4±2i}{10} when ± is plus. Add 4 to 2i.

$$x=\frac{4+2i}{10}$$

Divide 4+2i by 10.

$$x=\frac{2}{5}+\frac{1}{5}i$$

Now solve the equation x=\frac{4±2i}{10} when ± is minus. Subtract 2i from 4.

$$x=\frac{4-2i}{10}$$

Divide 4-2i by 10.

$$x=\frac{2}{5}-\frac{1}{5}i$$

The equation is now solved.

$$x=\frac{2}{5}+\frac{1}{5}i x=\frac{2}{5}-\frac{1}{5}i$$

Steps for Completing the Square

Subtract 4x from both sides.

$$5x^{2}-4x=-1$$

Divide both sides by 5.

$$\frac{5x^{2}-4x}{5}=-\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{4}{5}x=-\frac{1}{5}$$

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

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-\frac{1}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{1}{5}+\frac{4}{25}$$

Add -\frac{1}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{1}{25}$$

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

$$\left(x-\frac{2}{5}\right)^{2}=-\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{-\frac{1}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{1}{5}i x-\frac{2}{5}=-\frac{1}{5}i$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{2}{5}+\frac{1}{5}i x=\frac{2}{5}-\frac{1}{5}i$$