Solve 3+4a=-a^2

$$3 + 4 a = - a ^ { 2 }$$

Solve for a

$$a=-3$$

$$a=-1$$

Steps Using Factoring

Add a^{2} to both sides.

$$3+4a+a^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{2}+4a+3=0$$

To solve the equation, factor a^{2}+4a+3 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a+1\right)\left(a+3\right)$$

To find equation solutions, solve a+1=0 and a+3=0.

$$a=-1 a=-3$$

Steps Using Factoring By Grouping

Add a^{2} to both sides.

$$3+4a+a^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{2}+4a+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite a^{2}+4a+3 as \left(a^{2}+a\right)+\left(3a+3\right).

$$\left(a^{2}+a\right)+\left(3a+3\right)$$

Factor out a in the first and 3 in the second group.

$$a\left(a+1\right)+3\left(a+1\right)$$

Factor out common term a+1 by using distributive property.

$$\left(a+1\right)\left(a+3\right)$$

To find equation solutions, solve a+1=0 and a+3=0.

$$a=-1 a=-3$$

Steps Using the Quadratic Formula

Add a^{2} to both sides.

$$3+4a+a^{2}=0$$

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.

$$a^{2}+4a+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

Square 4.

$$a=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Multiply -4 times 3.

$$a=\frac{-4±\sqrt{16-12}}{2}$$

Add 16 to -12.

$$a=\frac{-4±\sqrt{4}}{2}$$

Take the square root of 4.

$$a=\frac{-4±2}{2}$$

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

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

Divide -2 by 2.

$$a=-1$$

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

$$a=-\frac{6}{2}$$

Divide -6 by 2.

$$a=-3$$

The equation is now solved.

$$a=-1 a=-3$$

Steps for Completing the Square

Add a^{2} to both sides.

$$3+4a+a^{2}=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4a+a^{2}=-3$$

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.

$$a^{2}+4a=-3$$

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.

$$a^{2}+4a+2^{2}=-3+2^{2}$$

Square 2.

$$a^{2}+4a+4=-3+4$$

Add -3 to 4.

$$a^{2}+4a+4=1$$

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}.

$$\left(a+2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+2\right)^{2}}=\sqrt{1}$$

Simplify.

$$a+2=1 a+2=-1$$

Subtract 2 from both sides of the equation.

$$a=-1 a=-3$$