Solve 5x^2=6x+2

$$5 x ^ { 2 } = 6 x + 2$$

Solve for x

$$x = \frac{\sqrt{19} + 3}{5} \approx 1.471779789$$

$$x=\frac{3-\sqrt{19}}{5}\approx -0.271779789$$

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$5x^{2}-6x=2$$

Subtract 2 from both sides.

$$5x^{2}-6x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -6 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5\left(-2\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+40}}{2\times 5}$$

Add 36 to 40.

$$x=\frac{-\left(-6\right)±\sqrt{76}}{2\times 5}$$

Take the square root of 76.

$$x=\frac{-\left(-6\right)±2\sqrt{19}}{2\times 5}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{19}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{6±2\sqrt{19}}{10}$$

Now solve the equation x=\frac{6±2\sqrt{19}}{10} when ± is plus. Add 6 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}+6}{10}$$

Divide 6+2\sqrt{19} by 10.

$$x=\frac{\sqrt{19}+3}{5}$$

Now solve the equation x=\frac{6±2\sqrt{19}}{10} when ± is minus. Subtract 2\sqrt{19} from 6.

$$x=\frac{6-2\sqrt{19}}{10}$$

Divide 6-2\sqrt{19} by 10.

$$x=\frac{3-\sqrt{19}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+3}{5} x=\frac{3-\sqrt{19}}{5}$$

Steps for Completing the Square

Subtract 6x from both sides.

$$5x^{2}-6x=2$$

Divide both sides by 5.

$$\frac{5x^{2}-6x}{5}=\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{6}{5}x=\frac{2}{5}$$

Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=\frac{2}{5}+\left(-\frac{3}{5}\right)^{2}$$

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{2}{5}+\frac{9}{25}$$

Add \frac{2}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{19}{25}$$

Factor x^{2}-\frac{6}{5}x+\frac{9}{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}.

$$\left(x-\frac{3}{5}\right)^{2}=\frac{19}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{5}\right)^{2}}=\sqrt{\frac{19}{25}}$$

Simplify.

$$x-\frac{3}{5}=\frac{\sqrt{19}}{5} x-\frac{3}{5}=-\frac{\sqrt{19}}{5}$$

Add \frac{3}{5} to both sides of the equation.

$$x=\frac{\sqrt{19}+3}{5} x=\frac{3-\sqrt{19}}{5}$$