Solve a^2+2=a-4

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

Solve for a

$$a=\frac{1+\sqrt{23}i}{2}\approx 0.5+2.397915762i$$

$$a=\frac{-\sqrt{23}i+1}{2}\approx 0.5-2.397915762i$$

Steps Using the Quadratic Formula

Subtract a from both sides.

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

Add 4 to both sides.

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

Add 2 and 4 to get 6.

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

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.

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

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

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

Multiply -4 times 6.

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

Add 1 to -24.

$$a=\frac{-\left(-1\right)±\sqrt{-23}}{2}$$

Take the square root of -23.

$$a=\frac{-\left(-1\right)±\sqrt{23}i}{2}$$

The opposite of -1 is 1.

$$a=\frac{1±\sqrt{23}i}{2}$$

Now solve the equation a=\frac{1±\sqrt{23}i}{2} when ± is plus. Add 1 to i\sqrt{23}.

$$a=\frac{1+\sqrt{23}i}{2}$$

Now solve the equation a=\frac{1±\sqrt{23}i}{2} when ± is minus. Subtract i\sqrt{23} from 1.

$$a=\frac{-\sqrt{23}i+1}{2}$$

The equation is now solved.

$$a=\frac{1+\sqrt{23}i}{2} a=\frac{-\sqrt{23}i+1}{2}$$

Steps for Completing the Square

Subtract a from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from -4 to get -6.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-a+\frac{1}{4}=-6+\frac{1}{4}$$

Add -6 to \frac{1}{4}.

$$a^{2}-a+\frac{1}{4}=-\frac{23}{4}$$

Factor a^{2}-a+\frac{1}{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}.

$$\left(a-\frac{1}{2}\right)^{2}=-\frac{23}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}$$

Simplify.

$$a-\frac{1}{2}=\frac{\sqrt{23}i}{2} a-\frac{1}{2}=-\frac{\sqrt{23}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$a=\frac{1+\sqrt{23}i}{2} a=\frac{-\sqrt{23}i+1}{2}$$