Solve -4+a+5=2+a^2+6a+5

$$- 4 + a + 5 = 2 + a ^ { 2 } + 6 a + 5$$

Solve for a

$$a=-3$$

$$a=-2$$

Steps Using Factoring By Grouping

Add -4 and 5 to get 1.

$$1+a=2+a^{2}+6a+5$$

Add 2 and 5 to get 7.

$$1+a=7+a^{2}+6a$$

Subtract 7 from both sides.

$$1+a-7=a^{2}+6a$$

Subtract 7 from 1 to get -6.

$$-6+a=a^{2}+6a$$

Subtract a^{2} from both sides.

$$-6+a-a^{2}=6a$$

Subtract 6a from both sides.

$$-6+a-a^{2}-6a=0$$

Combine a and -6a to get -5a.

$$-6-5a-a^{2}=0$$

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

$$-a^{2}-5a-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-2 b=-3$$

Rewrite -a^{2}-5a-6 as \left(-a^{2}-2a\right)+\left(-3a-6\right).

$$\left(-a^{2}-2a\right)+\left(-3a-6\right)$$

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

$$a\left(-a-2\right)+3\left(-a-2\right)$$

Factor out common term -a-2 by using distributive property.

$$\left(-a-2\right)\left(a+3\right)$$

To find equation solutions, solve -a-2=0 and a+3=0.

$$a=-2 a=-3$$

Steps Using the Quadratic Formula

Add -4 and 5 to get 1.

$$1+a=2+a^{2}+6a+5$$

Add 2 and 5 to get 7.

$$1+a=7+a^{2}+6a$$

Subtract 7 from both sides.

$$1+a-7=a^{2}+6a$$

Subtract 7 from 1 to get -6.

$$-6+a=a^{2}+6a$$

Subtract a^{2} from both sides.

$$-6+a-a^{2}=6a$$

Subtract 6a from both sides.

$$-6+a-a^{2}-6a=0$$

Combine a and -6a to get -5a.

$$-6-5a-a^{2}=0$$

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.

$$-a^{2}-5a-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-\left(-5\right)±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$a=\frac{-\left(-5\right)±\sqrt{25-24}}{2\left(-1\right)}$$

Add 25 to -24.

$$a=\frac{-\left(-5\right)±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$a=\frac{-\left(-5\right)±1}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$a=\frac{5±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{5±1}{-2}$$

Now solve the equation a=\frac{5±1}{-2} when ± is plus. Add 5 to 1.

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

Divide 6 by -2.

$$a=-3$$

Now solve the equation a=\frac{5±1}{-2} when ± is minus. Subtract 1 from 5.

$$a=\frac{4}{-2}$$

Divide 4 by -2.

$$a=-2$$

The equation is now solved.

$$a=-3 a=-2$$

Steps for Completing the Square

Add -4 and 5 to get 1.

$$1+a=2+a^{2}+6a+5$$

Add 2 and 5 to get 7.

$$1+a=7+a^{2}+6a$$

Subtract a^{2} from both sides.

$$1+a-a^{2}=7+6a$$

Subtract 6a from both sides.

$$1+a-a^{2}-6a=7$$

Combine a and -6a to get -5a.

$$1-5a-a^{2}=7$$

Subtract 1 from both sides.

$$-5a-a^{2}=7-1$$

Subtract 1 from 7 to get 6.

$$-5a-a^{2}=6$$

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.

$$-a^{2}-5a=6$$

Divide both sides by -1.

$$\frac{-a^{2}-5a}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\left(-\frac{5}{-1}\right)a=\frac{6}{-1}$$

Divide -5 by -1.

$$a^{2}+5a=\frac{6}{-1}$$

Divide 6 by -1.

$$a^{2}+5a=-6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+5a+\left(\frac{5}{2}\right)^{2}=-6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+5a+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$a^{2}+5a+\frac{25}{4}=\frac{1}{4}$$

Factor a^{2}+5a+\frac{25}{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}.

$$\left(a+\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$a+\frac{5}{2}=\frac{1}{2} a+\frac{5}{2}=-\frac{1}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$a=-2 a=-3$$