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3x_{0}^{2}+x_{0}=3
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.
3x_{0}^{2}+x_{0}-3=3-3
Subtract 3 from both sides of the equation.
3x_{0}^{2}+x_{0}-3=0
Subtracting 3 from itself leaves 0.
x_{0}=\frac{-1±\sqrt{1^{2}-4\times 3\left(-3\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x_{0}=\frac{-1±\sqrt{1-4\times 3\left(-3\right)}}{2\times 3}
Square 1.
x_{0}=\frac{-1±\sqrt{1-12\left(-3\right)}}{2\times 3}
Multiply -4 times 3.
x_{0}=\frac{-1±\sqrt{1+36}}{2\times 3}
Multiply -12 times -3.
x_{0}=\frac{-1±\sqrt{37}}{2\times 3}
Add 1 to 36.
x_{0}=\frac{-1±\sqrt{37}}{6}
Multiply 2 times 3.
x_{0}=\frac{\sqrt{37}-1}{6}
Now solve the equation x_{0}=\frac{-1±\sqrt{37}}{6} when ± is plus. Add -1 to \sqrt{37}.
x_{0}=\frac{-\sqrt{37}-1}{6}
Now solve the equation x_{0}=\frac{-1±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from -1.
x_{0}=\frac{\sqrt{37}-1}{6} x_{0}=\frac{-\sqrt{37}-1}{6}
The equation is now solved.
3x_{0}^{2}+x_{0}=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.
\frac{3x_{0}^{2}+x_{0}}{3}=\frac{3}{3}
Divide both sides by 3.
x_{0}^{2}+\frac{1}{3}x_{0}=\frac{3}{3}
Dividing by 3 undoes the multiplication by 3.
x_{0}^{2}+\frac{1}{3}x_{0}=1
Divide 3 by 3.
x_{0}^{2}+\frac{1}{3}x_{0}+\left(\frac{1}{6}\right)^{2}=1+\left(\frac{1}{6}\right)^{2}
Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x_{0}^{2}+\frac{1}{3}x_{0}+\frac{1}{36}=1+\frac{1}{36}
Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x_{0}^{2}+\frac{1}{3}x_{0}+\frac{1}{36}=\frac{37}{36}
Add 1 to \frac{1}{36}.
\left(x_{0}+\frac{1}{6}\right)^{2}=\frac{37}{36}
Factor x_{0}^{2}+\frac{1}{3}x_{0}+\frac{1}{36}. 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_{0}+\frac{1}{6}\right)^{2}}=\sqrt{\frac{37}{36}}
Take the square root of both sides of the equation.
x_{0}+\frac{1}{6}=\frac{\sqrt{37}}{6} x_{0}+\frac{1}{6}=-\frac{\sqrt{37}}{6}
Simplify.
x_{0}=\frac{\sqrt{37}-1}{6} x_{0}=\frac{-\sqrt{37}-1}{6}
Subtract \frac{1}{6} from both sides of the equation.