Solve 5x^2=4x+2

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

Solve for x

$$x = \frac{\sqrt{14} + 2}{5} \approx 1.148331477$$

$$x=\frac{2-\sqrt{14}}{5}\approx -0.348331477$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

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

Subtract 2 from both sides.

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

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

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

Square -4.

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

Multiply -4 times 5.

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

Multiply -20 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+40}}{2\times 5}$$

Add 16 to 40.

$$x=\frac{-\left(-4\right)±\sqrt{56}}{2\times 5}$$

Take the square root of 56.

$$x=\frac{-\left(-4\right)±2\sqrt{14}}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{14}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±2\sqrt{14}}{10}$$

Now solve the equation x=\frac{4±2\sqrt{14}}{10} when ± is plus. Add 4 to 2\sqrt{14}.

$$x=\frac{2\sqrt{14}+4}{10}$$

Divide 4+2\sqrt{14} by 10.

$$x=\frac{\sqrt{14}+2}{5}$$

Now solve the equation x=\frac{4±2\sqrt{14}}{10} when ± is minus. Subtract 2\sqrt{14} from 4.

$$x=\frac{4-2\sqrt{14}}{10}$$

Divide 4-2\sqrt{14} by 10.

$$x=\frac{2-\sqrt{14}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{14}+2}{5} x=\frac{2-\sqrt{14}}{5}$$

Steps for Completing the Square

Subtract 4x from both sides.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

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.

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

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

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

Add \frac{2}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{14}{25}$$

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}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{14}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{\sqrt{14}}{5} x-\frac{2}{5}=-\frac{\sqrt{14}}{5}$$

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

$$x=\frac{\sqrt{14}+2}{5} x=\frac{2-\sqrt{14}}{5}$$