Solve a=-a^2-4a+5+a-5

$$a = - a ^ { 2 } - 4 a + 5 + a - 5$$

Solve for a

$$a=-4$$

$$a=0$$

Steps Using Factoring

Combine -4a and a to get -3a.

$$a=-a^{2}-3a+5-5$$

Subtract 5 from 5 to get 0.

$$a=-a^{2}-3a$$

Add a^{2} to both sides.

$$a+a^{2}=-3a$$

Add 3a to both sides.

$$a+a^{2}+3a=0$$

Combine a and 3a to get 4a.

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

Factor out a.

$$a\left(4+a\right)=0$$

To find equation solutions, solve a=0 and 4+a=0.

$$a=0 a=-4$$

Steps Using the Quadratic Formula

Combine -4a and a to get -3a.

$$a=-a^{2}-3a+5-5$$

Subtract 5 from 5 to get 0.

$$a=-a^{2}-3a$$

Add a^{2} to both sides.

$$a+a^{2}=-3a$$

Add 3a to both sides.

$$a+a^{2}+3a=0$$

Combine a and 3a to get 4a.

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

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.

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

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

$$a=\frac{-4±\sqrt{4^{2}}}{2}$$

Take the square root of 4^{2}.

$$a=\frac{-4±4}{2}$$

Now solve the equation a=\frac{-4±4}{2} when ± is plus. Add -4 to 4.

$$a=\frac{0}{2}$$

Divide 0 by 2.

$$a=0$$

Now solve the equation a=\frac{-4±4}{2} when ± is minus. Subtract 4 from -4.

$$a=-\frac{8}{2}$$

Divide -8 by 2.

$$a=-4$$

The equation is now solved.

$$a=0 a=-4$$

Steps for Completing the Square

Combine -4a and a to get -3a.

$$a=-a^{2}-3a+5-5$$

Subtract 5 from 5 to get 0.

$$a=-a^{2}-3a$$

Add a^{2} to both sides.

$$a+a^{2}=-3a$$

Add 3a to both sides.

$$a+a^{2}+3a=0$$

Combine a and 3a to get 4a.

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

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.

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

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.

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

Square 2.

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

Factor a^{2}+4a+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$a+2=2 a+2=-2$$

Subtract 2 from both sides of the equation.

$$a=0 a=-4$$