Solve 5x^2=-40x

$$5 { x }^{ 2 } =-40x$$

Solve for x

$$x=-8$$

$$x=0$$

Steps Using Factoring

Add 40x to both sides.

$$5x^{2}+40x=0$$

Factor out x.

$$x\left(5x+40\right)=0$$

To find equation solutions, solve x=0 and 5x+40=0.

$$x=0 x=-8$$

Steps Using the Quadratic Formula

Add 40x to both sides.

$$5x^{2}+40x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 40^{2}.

$$x=\frac{-40±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-40±40}{10}$$

Now solve the equation x=\frac{-40±40}{10} when ± is plus. Add -40 to 40.

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

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-40±40}{10} when ± is minus. Subtract 40 from -40.

$$x=-\frac{80}{10}$$

Divide -80 by 10.

$$x=-8$$

The equation is now solved.

$$x=0 x=-8$$

Steps for Completing the Square

Add 40x to both sides.

$$5x^{2}+40x=0$$

Divide both sides by 5.

$$\frac{5x^{2}+40x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{40}{5}x=\frac{0}{5}$$

Divide 40 by 5.

$$x^{2}+8x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}+8x=0$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=4^{2}$$

Square 4.

$$x^{2}+8x+16=16$$

Factor x^{2}+8x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=4 x+4=-4$$

Subtract 4 from both sides of the equation.

$$x=0 x=-8$$