Solve (6-sqrt{3}y)^2+y^2=4

$$( 6 - \sqrt { 3 } y ) ^ { 2 } + y ^ { 2 } = 4$$

Solve for y

$$y=\frac{3\sqrt{3}+\sqrt{5}i}{2}\approx 2.598076211+1.118033989i$$

$$y=\frac{-\sqrt{5}i+3\sqrt{3}}{2}\approx 2.598076211-1.118033989i$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$\left(6-\sqrt{3}y\right)^{2}+y^{2}-4=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{3}y\right)^{2}.

$$36-12\sqrt{3}y+\left(\sqrt{3}\right)^{2}y^{2}+y^{2}-4=0$$

The square of \sqrt{3} is 3.

$$36-12\sqrt{3}y+3y^{2}+y^{2}-4=0$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$36-12\sqrt{3}y+4y^{2}-4=0$$

Subtract 4 from 36 to get 32.

$$32-12\sqrt{3}y+4y^{2}=0$$

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.

$$4y^{2}+\left(-12\sqrt{3}\right)y+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12\sqrt{3} for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\sqrt{3}\right)±\sqrt{\left(-12\sqrt{3}\right)^{2}-4\times 4\times 32}}{2\times 4}$$

Square -12\sqrt{3}.

$$y=\frac{-\left(-12\sqrt{3}\right)±\sqrt{432-4\times 4\times 32}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-12\sqrt{3}\right)±\sqrt{432-16\times 32}}{2\times 4}$$

Multiply -16 times 32.

$$y=\frac{-\left(-12\sqrt{3}\right)±\sqrt{432-512}}{2\times 4}$$

Add 432 to -512.

$$y=\frac{-\left(-12\sqrt{3}\right)±\sqrt{-80}}{2\times 4}$$

Take the square root of -80.

$$y=\frac{-\left(-12\sqrt{3}\right)±4\sqrt{5}i}{2\times 4}$$

The opposite of -12\sqrt{3} is 12\sqrt{3}.

$$y=\frac{12\sqrt{3}±4\sqrt{5}i}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{12\sqrt{3}±4\sqrt{5}i}{8}$$

Now solve the equation y=\frac{12\sqrt{3}±4\sqrt{5}i}{8} when ± is plus. Add 12\sqrt{3} to 4i\sqrt{5}.

$$y=\frac{12\sqrt{3}+4\sqrt{5}i}{8}$$

Divide 12\sqrt{3}+4i\sqrt{5} by 8.

$$y=\frac{3\sqrt{3}+\sqrt{5}i}{2}$$

Now solve the equation y=\frac{12\sqrt{3}±4\sqrt{5}i}{8} when ± is minus. Subtract 4i\sqrt{5} from 12\sqrt{3}.

$$y=\frac{-4\sqrt{5}i+12\sqrt{3}}{8}$$

Divide 12\sqrt{3}-4i\sqrt{5} by 8.

$$y=\frac{-\sqrt{5}i+3\sqrt{3}}{2}$$

The equation is now solved.

$$y=\frac{3\sqrt{3}+\sqrt{5}i}{2} y=\frac{-\sqrt{5}i+3\sqrt{3}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{3}y\right)^{2}.

$$36-12\sqrt{3}y+\left(\sqrt{3}\right)^{2}y^{2}+y^{2}=4$$

The square of \sqrt{3} is 3.

$$36-12\sqrt{3}y+3y^{2}+y^{2}=4$$

Combine 3y^{2} and y^{2} to get 4y^{2}.

$$36-12\sqrt{3}y+4y^{2}=4$$

Subtract 36 from both sides.

$$-12\sqrt{3}y+4y^{2}=4-36$$

Subtract 36 from 4 to get -32.

$$-12\sqrt{3}y+4y^{2}=-32$$

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.

$$4y^{2}+\left(-12\sqrt{3}\right)y=-32$$

Divide both sides by 4.

$$\frac{4y^{2}+\left(-12\sqrt{3}\right)y}{4}=-\frac{32}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\left(-\frac{12\sqrt{3}}{4}\right)y=-\frac{32}{4}$$

Divide -12\sqrt{3} by 4.

$$y^{2}+\left(-3\sqrt{3}\right)y=-\frac{32}{4}$$

Divide -32 by 4.

$$y^{2}+\left(-3\sqrt{3}\right)y=-8$$

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

$$y^{2}+\left(-3\sqrt{3}\right)y+\left(-\frac{3\sqrt{3}}{2}\right)^{2}=-8+\left(-\frac{3\sqrt{3}}{2}\right)^{2}$$

Square -\frac{3\sqrt{3}}{2}.

$$y^{2}+\left(-3\sqrt{3}\right)y+\frac{27}{4}=-8+\frac{27}{4}$$

Add -8 to \frac{27}{4}.

$$y^{2}+\left(-3\sqrt{3}\right)y+\frac{27}{4}=-\frac{5}{4}$$

Factor y^{2}+\left(-3\sqrt{3}\right)y+\frac{27}{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}.

$$\left(y-\frac{3\sqrt{3}}{2}\right)^{2}=-\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3\sqrt{3}}{2}\right)^{2}}=\sqrt{-\frac{5}{4}}$$

Simplify.

$$y-\frac{3\sqrt{3}}{2}=\frac{\sqrt{5}i}{2} y-\frac{3\sqrt{3}}{2}=-\frac{\sqrt{5}i}{2}$$

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

$$y=\frac{3\sqrt{3}+\sqrt{5}i}{2} y=\frac{-\sqrt{5}i+3\sqrt{3}}{2}$$