Solve 0=5x^2+20x

$$0=5 { x }^{ 2 } +20x$$

Solve for x

$$x=-4$$

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$5x^{2}+20x=0$$

Factor out x.

$$x\left(5x+20\right)=0$$

To find equation solutions, solve x=0 and 5x+20=0.

$$x=0 x=-4$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$5x^{2}+20x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 20^{2}.

$$x=\frac{-20±20}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-20±20}{10}$$

Now solve the equation x=\frac{-20±20}{10} when ± is plus. Add -20 to 20.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-20±20}{10} when ± is minus. Subtract 20 from -20.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

The equation is now solved.

$$x=0 x=-4$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$5x^{2}+20x=0$$

Divide both sides by 5.

$$\frac{5x^{2}+20x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{20}{5}x=\frac{0}{5}$$

Divide 20 by 5.

$$x^{2}+4x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}+4x=0$$

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

$$x^{2}+4x+2^{2}=2^{2}$$

Square 2.

$$x^{2}+4x+4=4$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+2=2 x+2=-2$$

Subtract 2 from both sides of the equation.

$$x=0 x=-4$$