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3x^{2}-2x=14
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^{2}-2x-14=14-14
Subtract 14 from both sides of the equation.
3x^{2}-2x-14=0
Subtracting 14 from itself leaves 0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-14\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-14\right)}}{2\times 3}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-12\left(-14\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-2\right)±\sqrt{4+168}}{2\times 3}
Multiply -12 times -14.
x=\frac{-\left(-2\right)±\sqrt{172}}{2\times 3}
Add 4 to 168.
x=\frac{-\left(-2\right)±2\sqrt{43}}{2\times 3}
Take the square root of 172.
x=\frac{2±2\sqrt{43}}{2\times 3}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{43}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{43}+2}{6}
Now solve the equation x=\frac{2±2\sqrt{43}}{6} when ± is plus. Add 2 to 2\sqrt{43}.
x=\frac{\sqrt{43}+1}{3}
Divide 2+2\sqrt{43} by 6.
x=\frac{2-2\sqrt{43}}{6}
Now solve the equation x=\frac{2±2\sqrt{43}}{6} when ± is minus. Subtract 2\sqrt{43} from 2.
x=\frac{1-\sqrt{43}}{3}
Divide 2-2\sqrt{43} by 6.
x=\frac{\sqrt{43}+1}{3} x=\frac{1-\sqrt{43}}{3}
The equation is now solved.
3x^{2}-2x=14
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^{2}-2x}{3}=\frac{14}{3}
Divide both sides by 3.
x^{2}-\frac{2}{3}x=\frac{14}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{14}{3}+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{14}{3}+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{43}{9}
Add \frac{14}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{3}\right)^{2}=\frac{43}{9}
Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. 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{1}{3}\right)^{2}}=\sqrt{\frac{43}{9}}
Take the square root of both sides of the equation.
x-\frac{1}{3}=\frac{\sqrt{43}}{3} x-\frac{1}{3}=-\frac{\sqrt{43}}{3}
Simplify.
x=\frac{\sqrt{43}+1}{3} x=\frac{1-\sqrt{43}}{3}
Add \frac{1}{3} to both sides of the equation.