Solve 10x^2=2x+3

$$10 x ^ { 2 } = 2 x + 3$$

Solve for x

$$x=\frac{\sqrt{31}+1}{10}\approx 0.656776436$$

$$x=\frac{1-\sqrt{31}}{10}\approx -0.456776436$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$10x^{2}-2x=3$$

Subtract 3 from both sides.

$$10x^{2}-2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-2\right)±\sqrt{4-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+120}}{2\times 10}$$

Add 4 to 120.

$$x=\frac{-\left(-2\right)±\sqrt{124}}{2\times 10}$$

Take the square root of 124.

$$x=\frac{-\left(-2\right)±2\sqrt{31}}{2\times 10}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{31}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{2±2\sqrt{31}}{20}$$

Now solve the equation x=\frac{2±2\sqrt{31}}{20} when ± is plus. Add 2 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}+2}{20}$$

Divide 2+2\sqrt{31} by 20.

$$x=\frac{\sqrt{31}+1}{10}$$

Now solve the equation x=\frac{2±2\sqrt{31}}{20} when ± is minus. Subtract 2\sqrt{31} from 2.

$$x=\frac{2-2\sqrt{31}}{20}$$

Divide 2-2\sqrt{31} by 20.

$$x=\frac{1-\sqrt{31}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{31}+1}{10} x=\frac{1-\sqrt{31}}{10}$$

Steps for Completing the Square

Subtract 2x from both sides.

$$10x^{2}-2x=3$$

Divide both sides by 10.

$$\frac{10x^{2}-2x}{10}=\frac{3}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{2}{10}\right)x=\frac{3}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{5}x=\frac{3}{10}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{3}{10}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{3}{10}+\frac{1}{100}$$

Add \frac{3}{10} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{31}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{31}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{31}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{\sqrt{31}}{10} x-\frac{1}{10}=-\frac{\sqrt{31}}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{\sqrt{31}+1}{10} x=\frac{1-\sqrt{31}}{10}$$