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

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

Solve for x

$$x=-2$$

$$x=-\frac{1}{5}=-0.2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by x+2.

$$5x^{2}+10x+x=-2$$

Combine 10x and x to get 11x.

$$5x^{2}+11x=-2$$

Add 2 to both sides.

$$5x^{2}+11x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 11 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 5\times 2}}{2\times 5}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 5\times 2}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-11±\sqrt{121-20\times 2}}{2\times 5}$$

Multiply -20 times 2.

$$x=\frac{-11±\sqrt{121-40}}{2\times 5}$$

Add 121 to -40.

$$x=\frac{-11±\sqrt{81}}{2\times 5}$$

Take the square root of 81.

$$x=\frac{-11±9}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-11±9}{10}$$

Now solve the equation x=\frac{-11±9}{10} when ± is plus. Add -11 to 9.

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

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

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

Now solve the equation x=\frac{-11±9}{10} when ± is minus. Subtract 9 from -11.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 5x by x+2.

$$5x^{2}+10x+x=-2$$

Combine 10x and x to get 11x.

$$5x^{2}+11x=-2$$

Divide both sides by 5.

$$\frac{5x^{2}+11x}{5}=-\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{11}{5}x=-\frac{2}{5}$$

Divide \frac{11}{5}, the coefficient of the x term, by 2 to get \frac{11}{10}. Then add the square of \frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{5}x+\left(\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(\frac{11}{10}\right)^{2}$$

Square \frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Add -\frac{2}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

Factor x^{2}+\frac{11}{5}x+\frac{121}{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}.

$$\left(x+\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x+\frac{11}{10}=\frac{9}{10} x+\frac{11}{10}=-\frac{9}{10}$$

Subtract \frac{11}{10} from both sides of the equation.

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