Solve 7x^2=4x-1

$$7 x ^ { 2 } = 4 x - 1$$

Solve for x (complex solution)

$$x=\frac{2+\sqrt{3}i}{7}\approx 0.285714286+0.24743583i$$

$$x=\frac{-\sqrt{3}i+2}{7}\approx 0.285714286-0.24743583i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$7x^{2}-4x=-1$$

Add 1 to both sides.

$$7x^{2}-4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 7}}{2\times 7}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 7}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-4\right)±\sqrt{16-28}}{2\times 7}$$

Add 16 to -28.

$$x=\frac{-\left(-4\right)±\sqrt{-12}}{2\times 7}$$

Take the square root of -12.

$$x=\frac{-\left(-4\right)±2\sqrt{3}i}{2\times 7}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{3}i}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{4±2\sqrt{3}i}{14}$$

Now solve the equation x=\frac{4±2\sqrt{3}i}{14} when ± is plus. Add 4 to 2i\sqrt{3}.

$$x=\frac{4+2\sqrt{3}i}{14}$$

Divide 4+2i\sqrt{3} by 14.

$$x=\frac{2+\sqrt{3}i}{7}$$

Now solve the equation x=\frac{4±2\sqrt{3}i}{14} when ± is minus. Subtract 2i\sqrt{3} from 4.

$$x=\frac{-2\sqrt{3}i+4}{14}$$

Divide 4-2i\sqrt{3} by 14.

$$x=\frac{-\sqrt{3}i+2}{7}$$

The equation is now solved.

$$x=\frac{2+\sqrt{3}i}{7} x=\frac{-\sqrt{3}i+2}{7}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$7x^{2}-4x=-1$$

Divide both sides by 7.

$$\frac{7x^{2}-4x}{7}=-\frac{1}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{4}{7}x=-\frac{1}{7}$$

Divide -\frac{4}{7}, the coefficient of the x term, by 2 to get -\frac{2}{7}. Then add the square of -\frac{2}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{7}x+\left(-\frac{2}{7}\right)^{2}=-\frac{1}{7}+\left(-\frac{2}{7}\right)^{2}$$

Square -\frac{2}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=-\frac{1}{7}+\frac{4}{49}$$

Add -\frac{1}{7} to \frac{4}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=-\frac{3}{49}$$

Factor x^{2}-\frac{4}{7}x+\frac{4}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{7}\right)^{2}=-\frac{3}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{7}\right)^{2}}=\sqrt{-\frac{3}{49}}$$

Simplify.

$$x-\frac{2}{7}=\frac{\sqrt{3}i}{7} x-\frac{2}{7}=-\frac{\sqrt{3}i}{7}$$

Add \frac{2}{7} to both sides of the equation.

$$x=\frac{2+\sqrt{3}i}{7} x=\frac{-\sqrt{3}i+2}{7}$$