Solve -1=-1/2(3+a)^2+2

$$- 1 = - \frac { 1 } { 2 } ( 3 + a ) ^ { 2 } + 2$$

Solve for a

$$a=\sqrt{6}-3\approx -0.550510257$$

$$a=-\left(\sqrt{6}+3\right)\approx -5.449489743$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+a\right)^{2}.

$$-1=-\frac{1}{2}\left(9+6a+a^{2}\right)+2$$

Use the distributive property to multiply -\frac{1}{2} by 9+6a+a^{2}.

$$-1=-\frac{9}{2}-3a-\frac{1}{2}a^{2}+2$$

Add -\frac{9}{2} and 2 to get -\frac{5}{2}.

$$-1=-\frac{5}{2}-3a-\frac{1}{2}a^{2}$$

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

$$-\frac{5}{2}-3a-\frac{1}{2}a^{2}=-1$$

Add 1 to both sides.

$$-\frac{5}{2}-3a-\frac{1}{2}a^{2}+1=0$$

Add -\frac{5}{2} and 1 to get -\frac{3}{2}.

$$-\frac{3}{2}-3a-\frac{1}{2}a^{2}=0$$

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.

$$-\frac{1}{2}a^{2}-3a-\frac{3}{2}=0$$

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

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}}{2\left(-\frac{1}{2}\right)}$$

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$a=\frac{-\left(-3\right)±\sqrt{9+2\left(-\frac{3}{2}\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{3}{2}.

$$a=\frac{-\left(-3\right)±\sqrt{9-3}}{2\left(-\frac{1}{2}\right)}$$

Add 9 to -3.

$$a=\frac{-\left(-3\right)±\sqrt{6}}{2\left(-\frac{1}{2}\right)}$$

The opposite of -3 is 3.

$$a=\frac{3±\sqrt{6}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$a=\frac{3±\sqrt{6}}{-1}$$

Now solve the equation a=\frac{3±\sqrt{6}}{-1} when ± is plus. Add 3 to \sqrt{6}.

$$a=\frac{\sqrt{6}+3}{-1}$$

Divide 3+\sqrt{6} by -1.

$$a=-\left(\sqrt{6}+3\right)$$

Now solve the equation a=\frac{3±\sqrt{6}}{-1} when ± is minus. Subtract \sqrt{6} from 3.

$$a=\frac{3-\sqrt{6}}{-1}$$

Divide 3-\sqrt{6} by -1.

$$a=\sqrt{6}-3$$

The equation is now solved.

$$a=-\left(\sqrt{6}+3\right) a=\sqrt{6}-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+a\right)^{2}.

$$-1=-\frac{1}{2}\left(9+6a+a^{2}\right)+2$$

Use the distributive property to multiply -\frac{1}{2} by 9+6a+a^{2}.

$$-1=-\frac{9}{2}-3a-\frac{1}{2}a^{2}+2$$

Add -\frac{9}{2} and 2 to get -\frac{5}{2}.

$$-1=-\frac{5}{2}-3a-\frac{1}{2}a^{2}$$

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

$$-\frac{5}{2}-3a-\frac{1}{2}a^{2}=-1$$

Add \frac{5}{2} to both sides.

$$-3a-\frac{1}{2}a^{2}=-1+\frac{5}{2}$$

Add -1 and \frac{5}{2} to get \frac{3}{2}.

$$-3a-\frac{1}{2}a^{2}=\frac{3}{2}$$

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.

$$-\frac{1}{2}a^{2}-3a=\frac{3}{2}$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}a^{2}-3a}{-\frac{1}{2}}=\frac{\frac{3}{2}}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$a^{2}+\left(-\frac{3}{-\frac{1}{2}}\right)a=\frac{\frac{3}{2}}{-\frac{1}{2}}$$

Divide -3 by -\frac{1}{2} by multiplying -3 by the reciprocal of -\frac{1}{2}.

$$a^{2}+6a=\frac{\frac{3}{2}}{-\frac{1}{2}}$$

Divide \frac{3}{2} by -\frac{1}{2} by multiplying \frac{3}{2} by the reciprocal of -\frac{1}{2}.

$$a^{2}+6a=-3$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+6a+3^{2}=-3+3^{2}$$

Square 3.

$$a^{2}+6a+9=-3+9$$

Add -3 to 9.

$$a^{2}+6a+9=6$$

Factor a^{2}+6a+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$a+3=\sqrt{6} a+3=-\sqrt{6}$$

Subtract 3 from both sides of the equation.

$$a=\sqrt{6}-3 a=-\sqrt{6}-3$$