Solve (5x-1)(2x+1)=2x

$$( 5 x - 1 ) ( 2 x + 1 ) = 2 x$$

Solve for x

$$x=\frac{\sqrt{41}-1}{20}\approx 0.270156212$$

$$x=\frac{-\sqrt{41}-1}{20}\approx -0.370156212$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x-1 by 2x+1 and combine like terms.

$$10x^{2}+3x-1=2x$$

Subtract 2x from both sides.

$$10x^{2}+3x-1-2x=0$$

Combine 3x and -2x to get x.

$$10x^{2}+x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 10\left(-1\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-1±\sqrt{1-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$x=\frac{-1±\sqrt{1+40}}{2\times 10}$$

Add 1 to 40.

$$x=\frac{-1±\sqrt{41}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-1±\sqrt{41}}{20}$$

Now solve the equation x=\frac{-1±\sqrt{41}}{20} when ± is plus. Add -1 to \sqrt{41}.

$$x=\frac{\sqrt{41}-1}{20}$$

Now solve the equation x=\frac{-1±\sqrt{41}}{20} when ± is minus. Subtract \sqrt{41} from -1.

$$x=\frac{-\sqrt{41}-1}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-1}{20} x=\frac{-\sqrt{41}-1}{20}$$

Steps for Completing the Square

Use the distributive property to multiply 5x-1 by 2x+1 and combine like terms.

$$10x^{2}+3x-1=2x$$

Subtract 2x from both sides.

$$10x^{2}+3x-1-2x=0$$

Combine 3x and -2x to get x.

$$10x^{2}+x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$10x^{2}+x=1$$

Divide both sides by 10.

$$\frac{10x^{2}+x}{10}=\frac{1}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{1}{10}x=\frac{1}{10}$$

Divide \frac{1}{10}, the coefficient of the x term, by 2 to get \frac{1}{20}. Then add the square of \frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{10}x+\left(\frac{1}{20}\right)^{2}=\frac{1}{10}+\left(\frac{1}{20}\right)^{2}$$

Square \frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{1}{10}+\frac{1}{400}$$

Add \frac{1}{10} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{41}{400}$$

Factor x^{2}+\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{20}\right)^{2}=\frac{41}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{20}\right)^{2}}=\sqrt{\frac{41}{400}}$$

Simplify.

$$x+\frac{1}{20}=\frac{\sqrt{41}}{20} x+\frac{1}{20}=-\frac{\sqrt{41}}{20}$$

Subtract \frac{1}{20} from both sides of the equation.

$$x=\frac{\sqrt{41}-1}{20} x=\frac{-\sqrt{41}-1}{20}$$