Solve =pi/2quadx^2-(3-sqrt{3})x+2-sqrt{3}=0

$$= \frac { \pi } { 2 } \quad x ^ { 2 } - ( 3 - \sqrt { 3 } ) x + 2 - \sqrt { 3 } = 0$$

Solve for x (complex solution)

$$x=\frac{3-\sqrt{3}+i\sqrt{2\left(-\pi \sqrt{3}+2\pi +3\sqrt{3}-6\right)}}{\pi }\approx 0.403600763+0.087682255i$$

$$x=\frac{-i\sqrt{2\left(-\pi \sqrt{3}+2\pi +3\sqrt{3}-6\right)}+3-\sqrt{3}}{\pi }\approx 0.403600763-0.087682255i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2\left(\frac{\pi }{2}x^{2}-\left(3-\sqrt{3}\right)x\right)+4-2\sqrt{3}=0$$

Express \frac{\pi }{2}x^{2} as a single fraction.

$$2\left(\frac{\pi x^{2}}{2}-\left(3-\sqrt{3}\right)x\right)+4-2\sqrt{3}=0$$

Use the distributive property to multiply 3-\sqrt{3} by x.

$$2\left(\frac{\pi x^{2}}{2}-\left(3x-\sqrt{3}x\right)\right)+4-2\sqrt{3}=0$$

To find the opposite of 3x-\sqrt{3}x, find the opposite of each term.

$$2\left(\frac{\pi x^{2}}{2}-3x+\sqrt{3}x\right)+4-2\sqrt{3}=0$$

Use the distributive property to multiply 2 by \frac{\pi x^{2}}{2}-3x+\sqrt{3}x.

$$2\times \frac{\pi x^{2}}{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Express 2\times \frac{\pi x^{2}}{2} as a single fraction.

$$\frac{2\pi x^{2}}{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Cancel out 2 and 2.

$$\pi x^{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Combine all terms containing x.

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

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.

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

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

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

Square -6+2\sqrt{3}.

$$x=\frac{-\left(2\sqrt{3}-6\right)±\sqrt{48-24\sqrt{3}-4\pi \left(4-2\sqrt{3}\right)}}{2\pi }$$

Multiply -4 times \pi .

$$x=\frac{-\left(2\sqrt{3}-6\right)±\sqrt{48-24\sqrt{3}+\left(-4\pi \right)\left(4-2\sqrt{3}\right)}}{2\pi }$$

Multiply -4\pi times 4-2\sqrt{3}.

$$x=\frac{-\left(2\sqrt{3}-6\right)±\sqrt{48-24\sqrt{3}-8\pi \left(2-\sqrt{3}\right)}}{2\pi }$$

Add 48-24\sqrt{3} to -8\pi \left(2-\sqrt{3}\right).

$$x=\frac{-\left(2\sqrt{3}-6\right)±\sqrt{8\pi \sqrt{3}+48-16\pi -24\sqrt{3}}}{2\pi }$$

Take the square root of 48-24\sqrt{3}-16\pi +8\pi \sqrt{3}.

$$x=\frac{-\left(2\sqrt{3}-6\right)±2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{2\pi }$$

The opposite of -6+2\sqrt{3} is 6-2\sqrt{3}.

$$x=\frac{6-2\sqrt{3}±2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{2\pi }$$

Now solve the equation x=\frac{6-2\sqrt{3}±2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{2\pi } when ± is plus. Add 6-2\sqrt{3} to 2i\sqrt{-12+6\sqrt{3}+4\pi -2\pi \sqrt{3}}.

$$x=\frac{6-2\sqrt{3}+2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{2\pi }$$

Divide 6-2\sqrt{3}+2i\sqrt{-12+6\sqrt{3}+4\pi -2\sqrt{3}\pi } by 2\pi .

$$x=\frac{3-\sqrt{3}+i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{\pi }$$

Now solve the equation x=\frac{6-2\sqrt{3}±2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{2\pi } when ± is minus. Subtract 2i\sqrt{-12+6\sqrt{3}+4\pi -2\pi \sqrt{3}} from 6-2\sqrt{3}.

$$x=\frac{-2i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}+6-2\sqrt{3}}{2\pi }$$

Divide 6-2\sqrt{3}-2i\sqrt{-12+6\sqrt{3}+4\pi -2\sqrt{3}\pi } by 2\pi .

$$x=\frac{-i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}+3-\sqrt{3}}{\pi }$$

The equation is now solved.

$$x=\frac{3-\sqrt{3}+i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{\pi } x=\frac{-i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}+3-\sqrt{3}}{\pi }$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2\left(\frac{\pi }{2}x^{2}-\left(3-\sqrt{3}\right)x\right)+4-2\sqrt{3}=0$$

Express \frac{\pi }{2}x^{2} as a single fraction.

$$2\left(\frac{\pi x^{2}}{2}-\left(3-\sqrt{3}\right)x\right)+4-2\sqrt{3}=0$$

Use the distributive property to multiply 3-\sqrt{3} by x.

$$2\left(\frac{\pi x^{2}}{2}-\left(3x-\sqrt{3}x\right)\right)+4-2\sqrt{3}=0$$

To find the opposite of 3x-\sqrt{3}x, find the opposite of each term.

$$2\left(\frac{\pi x^{2}}{2}-3x+\sqrt{3}x\right)+4-2\sqrt{3}=0$$

Use the distributive property to multiply 2 by \frac{\pi x^{2}}{2}-3x+\sqrt{3}x.

$$2\times \frac{\pi x^{2}}{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Express 2\times \frac{\pi x^{2}}{2} as a single fraction.

$$\frac{2\pi x^{2}}{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Cancel out 2 and 2.

$$\pi x^{2}-6x+2\sqrt{3}x+4-2\sqrt{3}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$\pi x^{2}-6x+2\sqrt{3}x-2\sqrt{3}=-4$$

Add 2\sqrt{3} to both sides.

$$\pi x^{2}-6x+2\sqrt{3}x=-4+2\sqrt{3}$$

Combine all terms containing x.

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

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.

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

Divide both sides by \pi .

$$\frac{\pi x^{2}+\left(2\sqrt{3}-6\right)x}{\pi }=\frac{2\sqrt{3}-4}{\pi }$$

Dividing by \pi undoes the multiplication by \pi .

$$x^{2}+\frac{2\sqrt{3}-6}{\pi }x=\frac{2\sqrt{3}-4}{\pi }$$

Divide -6+2\sqrt{3} by \pi .

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

Divide -4+2\sqrt{3} by \pi .

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

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

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

Square \frac{-3+\sqrt{3}}{\pi }.

$$x^{2}+\frac{2\left(\sqrt{3}-3\right)}{\pi }x+\frac{12-6\sqrt{3}}{\pi ^{2}}=\frac{2\left(\sqrt{3}-2\right)}{\pi }+\frac{12-6\sqrt{3}}{\pi ^{2}}$$

Add \frac{2\left(-2+\sqrt{3}\right)}{\pi } to \frac{12-6\sqrt{3}}{\pi ^{2}}.

$$x^{2}+\frac{2\left(\sqrt{3}-3\right)}{\pi }x+\frac{12-6\sqrt{3}}{\pi ^{2}}=\frac{2\left(\pi \sqrt{3}+6-2\pi -3\sqrt{3}\right)}{\pi ^{2}}$$

Factor x^{2}+\frac{2\left(\sqrt{3}-3\right)}{\pi }x+\frac{12-6\sqrt{3}}{\pi ^{2}}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{\sqrt{3}-3}{\pi }=\frac{i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{\pi } x+\frac{\sqrt{3}-3}{\pi }=-\frac{i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{\pi }$$

Subtract \frac{-3+\sqrt{3}}{\pi } from both sides of the equation.

$$x=\frac{3-\sqrt{3}+i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}}{\pi } x=\frac{-i\sqrt{-2\pi \sqrt{3}+4\pi +6\sqrt{3}-12}+3-\sqrt{3}}{\pi }$$