Solve 2x+11x^2+5x=0

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

Solve for x

$$x=-\frac{7}{11}\approx -0.636363636$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(2+11x+5\right)=0$$

To find equation solutions, solve x=0 and 7+11x=0.

$$x=0 x=-\frac{7}{11}$$

Steps Using the Quadratic Formula

Combine 2x and 5x to get 7x.

$$7x+11x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$11x^{2}+7x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, 7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}}}{2\times 11}$$

Take the square root of 7^{2}.

$$x=\frac{-7±7}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{-7±7}{22}$$

Now solve the equation x=\frac{-7±7}{22} when ± is plus. Add -7 to 7.

$$x=\frac{0}{22}$$

Divide 0 by 22.

$$x=0$$

Now solve the equation x=\frac{-7±7}{22} when ± is minus. Subtract 7 from -7.

$$x=-\frac{14}{22}$$

Reduce the fraction \frac{-14}{22} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{11}$$

The equation is now solved.

$$x=0 x=-\frac{7}{11}$$

Steps for Completing the Square

Combine 2x and 5x to get 7x.

$$7x+11x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$11x^{2}+7x=0$$

Divide both sides by 11.

$$\frac{11x^{2}+7x}{11}=\frac{0}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$x^{2}+\frac{7}{11}x=\frac{0}{11}$$

Divide 0 by 11.

$$x^{2}+\frac{7}{11}x=0$$

Divide \frac{7}{11}, the coefficient of the x term, by 2 to get \frac{7}{22}. Then add the square of \frac{7}{22} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{11}x+\left(\frac{7}{22}\right)^{2}=\left(\frac{7}{22}\right)^{2}$$

Square \frac{7}{22} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{11}x+\frac{49}{484}=\frac{49}{484}$$

Factor x^{2}+\frac{7}{11}x+\frac{49}{484}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{22}\right)^{2}=\frac{49}{484}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{22}\right)^{2}}=\sqrt{\frac{49}{484}}$$

Simplify.

$$x+\frac{7}{22}=\frac{7}{22} x+\frac{7}{22}=-\frac{7}{22}$$

Subtract \frac{7}{22} from both sides of the equation.

$$x=0 x=-\frac{7}{11}$$