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\sqrt{3}x^{2}-2x+\frac{3}{5}=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.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\sqrt{3}\times \frac{3}{5}}}{2\sqrt{3}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \sqrt{3} for a, -2 for b, and \frac{3}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\sqrt{3}\times \frac{3}{5}}}{2\sqrt{3}}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+\left(-4\sqrt{3}\right)\times \frac{3}{5}}}{2\sqrt{3}}
Multiply -4 times \sqrt{3}.
x=\frac{-\left(-2\right)±\sqrt{4-\frac{12\sqrt{3}}{5}}}{2\sqrt{3}}
Multiply -4\sqrt{3} times \frac{3}{5}.
x=\frac{-\left(-2\right)±\sqrt{-\frac{12\sqrt{3}}{5}+4}}{2\sqrt{3}}
Add 4 to -\frac{12\sqrt{3}}{5}.
x=\frac{-\left(-2\right)±\frac{2i\sqrt{15\sqrt{3}-25}}{5}}{2\sqrt{3}}
Take the square root of 4-\frac{12\sqrt{3}}{5}.
x=\frac{2±\frac{2i\sqrt{15\sqrt{3}-25}}{5}}{2\sqrt{3}}
The opposite of -2 is 2.
x=\frac{\frac{2i\sqrt{15\sqrt{3}-25}}{5}+2}{2\sqrt{3}}
Now solve the equation x=\frac{2±\frac{2i\sqrt{15\sqrt{3}-25}}{5}}{2\sqrt{3}} when ± is plus. Add 2 to \frac{2i\sqrt{-25+15\sqrt{3}}}{5}.
x=\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3}
Divide 2+\frac{2i\sqrt{-25+15\sqrt{3}}}{5} by 2\sqrt{3}.
x=\frac{-\frac{2i\sqrt{15\sqrt{3}-25}}{5}+2}{2\sqrt{3}}
Now solve the equation x=\frac{2±\frac{2i\sqrt{15\sqrt{3}-25}}{5}}{2\sqrt{3}} when ± is minus. Subtract \frac{2i\sqrt{-25+15\sqrt{3}}}{5} from 2.
x=-\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3}
Divide 2-\frac{2i\sqrt{-25+15\sqrt{3}}}{5} by 2\sqrt{3}.
x=\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3} x=-\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3}
The equation is now solved.
\sqrt{3}x^{2}-2x+\frac{3}{5}=0
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.
\sqrt{3}x^{2}-2x+\frac{3}{5}-\frac{3}{5}=-\frac{3}{5}
Subtract \frac{3}{5} from both sides of the equation.
\sqrt{3}x^{2}-2x=-\frac{3}{5}
Subtracting \frac{3}{5} from itself leaves 0.
\frac{\sqrt{3}x^{2}-2x}{\sqrt{3}}=-\frac{\frac{3}{5}}{\sqrt{3}}
Divide both sides by \sqrt{3}.
x^{2}+\left(-\frac{2}{\sqrt{3}}\right)x=-\frac{\frac{3}{5}}{\sqrt{3}}
Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.
x^{2}+\left(-\frac{2\sqrt{3}}{3}\right)x=-\frac{\frac{3}{5}}{\sqrt{3}}
Divide -2 by \sqrt{3}.
x^{2}+\left(-\frac{2\sqrt{3}}{3}\right)x=-\frac{\sqrt{3}}{5}
Divide -\frac{3}{5} by \sqrt{3}.
x^{2}+\left(-\frac{2\sqrt{3}}{3}\right)x+\left(-\frac{\sqrt{3}}{3}\right)^{2}=-\frac{\sqrt{3}}{5}+\left(-\frac{\sqrt{3}}{3}\right)^{2}
Divide -\frac{2\sqrt{3}}{3}, the coefficient of the x term, by 2 to get -\frac{\sqrt{3}}{3}. Then add the square of -\frac{\sqrt{3}}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\left(-\frac{2\sqrt{3}}{3}\right)x+\frac{1}{3}=-\frac{\sqrt{3}}{5}+\frac{1}{3}
Square -\frac{\sqrt{3}}{3}.
\left(x-\frac{\sqrt{3}}{3}\right)^{2}=-\frac{\sqrt{3}}{5}+\frac{1}{3}
Factor x^{2}+\left(-\frac{2\sqrt{3}}{3}\right)x+\frac{1}{3}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{\sqrt{3}}{3}\right)^{2}}=\sqrt{-\frac{\sqrt{3}}{5}+\frac{1}{3}}
Take the square root of both sides of the equation.
x-\frac{\sqrt{3}}{3}=\frac{i\sqrt{45\sqrt{3}-75}}{15} x-\frac{\sqrt{3}}{3}=-\frac{i\sqrt{45\sqrt{3}-75}}{15}
Simplify.
x=\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3} x=-\frac{i\sqrt{45\sqrt{3}-75}}{15}+\frac{\sqrt{3}}{3}
Add \frac{\sqrt{3}}{3} to both sides of the equation.