Solve y=y^2-3

$$y = y ^ { 2 } - 3$$

Solve for y

$$y = \frac{\sqrt{13} + 1}{2} \approx 2.302775638$$

$$y=\frac{1-\sqrt{13}}{2}\approx -1.302775638$$

Steps Using the Quadratic Formula

Subtract y^{2} from both sides.

$$y-y^{2}=-3$$

Add 3 to both sides.

$$y-y^{2}+3=0$$

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.

$$-y^{2}+y+3=0$$

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

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

Square 1.

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

Multiply -4 times -1.

$$y=\frac{-1±\sqrt{1+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$y=\frac{-1±\sqrt{1+12}}{2\left(-1\right)}$$

Add 1 to 12.

$$y=\frac{-1±\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-1±\sqrt{13}}{-2}$$

Now solve the equation y=\frac{-1±\sqrt{13}}{-2} when ± is plus. Add -1 to \sqrt{13}.

$$y=\frac{\sqrt{13}-1}{-2}$$

Divide -1+\sqrt{13} by -2.

$$y=\frac{1-\sqrt{13}}{2}$$

Now solve the equation y=\frac{-1±\sqrt{13}}{-2} when ± is minus. Subtract \sqrt{13} from -1.

$$y=\frac{-\sqrt{13}-1}{-2}$$

Divide -1-\sqrt{13} by -2.

$$y=\frac{\sqrt{13}+1}{2}$$

The equation is now solved.

$$y=\frac{1-\sqrt{13}}{2} y=\frac{\sqrt{13}+1}{2}$$

Steps for Completing the Square

Subtract y^{2} from both sides.

$$y-y^{2}=-3$$

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.

$$-y^{2}+y=-3$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -3 by -1.

$$y^{2}-y=3$$

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.

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

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

$$y^{2}-y+\frac{1}{4}=3+\frac{1}{4}$$

Add 3 to \frac{1}{4}.

$$y^{2}-y+\frac{1}{4}=\frac{13}{4}$$

Factor y^{2}-y+\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}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{\sqrt{13}}{2} y-\frac{1}{2}=-\frac{\sqrt{13}}{2}$$

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

$$y=\frac{\sqrt{13}+1}{2} y=\frac{1-\sqrt{13}}{2}$$