Solve x(2x+1)/x-2=10/x-2

$$\frac { x ( 2 x + 1 ) } { x - 2 } = \frac { 10 } { x - 2 }$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x\left(2x+1\right)=10$$

Use the distributive property to multiply x by 2x+1.

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

Subtract 10 from both sides.

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

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

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-1±\sqrt{1+80}}{2\times 2}$$

Add 1 to 80.

$$x=\frac{-1±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-1±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±9}{4}$$

Now solve the equation x=\frac{-1±9}{4} when ± is plus. Add -1 to 9.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-1±9}{4} when ± is minus. Subtract 9 from -1.

$$x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Variable x cannot be equal to 2.

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

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x\left(2x+1\right)=10$$

Use the distributive property to multiply x by 2x+1.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 10 by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=5+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=5+\frac{1}{16}$$

Add 5 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{9}{4} x+\frac{1}{4}=-\frac{9}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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

Variable x cannot be equal to 2.

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