Solve (3-y)^2+y^2=12

$$( 3 - y ) ^ { 2 } + y ^ { 2 } = 12$$

Solve for y

$$y = \frac{\sqrt{15} + 3}{2} \approx 3.436491673$$

$$y=\frac{3-\sqrt{15}}{2}\approx -0.436491673$$

Steps Using the Quadratic Formula

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

$$9-6y+y^{2}+y^{2}=12$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$9-6y+2y^{2}=12$$

Subtract 12 from both sides.

$$9-6y+2y^{2}-12=0$$

Subtract 12 from 9 to get -3.

$$-3-6y+2y^{2}=0$$

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.

$$2y^{2}-6y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-6\right)±\sqrt{36-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$y=\frac{-\left(-6\right)±\sqrt{36+24}}{2\times 2}$$

Add 36 to 24.

$$y=\frac{-\left(-6\right)±\sqrt{60}}{2\times 2}$$

Take the square root of 60.

$$y=\frac{-\left(-6\right)±2\sqrt{15}}{2\times 2}$$

The opposite of -6 is 6.

$$y=\frac{6±2\sqrt{15}}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{6±2\sqrt{15}}{4}$$

Now solve the equation y=\frac{6±2\sqrt{15}}{4} when ± is plus. Add 6 to 2\sqrt{15}.

$$y=\frac{2\sqrt{15}+6}{4}$$

Divide 6+2\sqrt{15} by 4.

$$y=\frac{\sqrt{15}+3}{2}$$

Now solve the equation y=\frac{6±2\sqrt{15}}{4} when ± is minus. Subtract 2\sqrt{15} from 6.

$$y=\frac{6-2\sqrt{15}}{4}$$

Divide 6-2\sqrt{15} by 4.

$$y=\frac{3-\sqrt{15}}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{15}+3}{2} y=\frac{3-\sqrt{15}}{2}$$

Steps for Completing the Square

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

$$9-6y+y^{2}+y^{2}=12$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$9-6y+2y^{2}=12$$

Subtract 9 from both sides.

$$-6y+2y^{2}=12-9$$

Subtract 9 from 12 to get 3.

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

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.

$$2y^{2}-6y=3$$

Divide both sides by 2.

$$\frac{2y^{2}-6y}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\left(-\frac{6}{2}\right)y=\frac{3}{2}$$

Divide -6 by 2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=\frac{3}{2}+\frac{9}{4}$$

Add \frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-3y+\frac{9}{4}=\frac{15}{4}$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{15}{4}}$$

Simplify.

$$y-\frac{3}{2}=\frac{\sqrt{15}}{2} y-\frac{3}{2}=-\frac{\sqrt{15}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$y=\frac{\sqrt{15}+3}{2} y=\frac{3-\sqrt{15}}{2}$$