Solve -a=-a^2-2a+3+1

$$- a = - a ^ { 2 } - 2 a + 3 + 1$$

Solve for a

$$a = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$$

$$a=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

Steps Using the Quadratic Formula

Add 3 and 1 to get 4.

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

Add a^{2} to both sides.

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

Add 2a to both sides.

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

Subtract 4 from both sides.

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

Combine -a and 2a to get a.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -4.

$$a=\frac{-1±\sqrt{1+16}}{2}$$

Add 1 to 16.

$$a=\frac{-1±\sqrt{17}}{2}$$

Now solve the equation a=\frac{-1±\sqrt{17}}{2} when ± is plus. Add -1 to \sqrt{17}.

$$a=\frac{\sqrt{17}-1}{2}$$

Now solve the equation a=\frac{-1±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from -1.

$$a=\frac{-\sqrt{17}-1}{2}$$

The equation is now solved.

$$a=\frac{\sqrt{17}-1}{2} a=\frac{-\sqrt{17}-1}{2}$$

Steps for Completing the Square

Add 3 and 1 to get 4.

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

Add a^{2} to both sides.

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

Add 2a to both sides.

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

Combine -a and 2a to get a.

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

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.

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

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.

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

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

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

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

$$a^{2}+a+\frac{1}{4}=\frac{17}{4}$$

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}.

$$\left(a+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$a+\frac{1}{2}=\frac{\sqrt{17}}{2} a+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$a=\frac{\sqrt{17}-1}{2} a=\frac{-\sqrt{17}-1}{2}$$