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3x^{2}-x=5
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}-x-5=5-5
Subtract 5 from both sides of the equation.
3x^{2}-x-5=0
Subtracting 5 from itself leaves 0.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-5\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-12\left(-5\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-1\right)±\sqrt{1+60}}{2\times 3}
Multiply -12 times -5.
x=\frac{-\left(-1\right)±\sqrt{61}}{2\times 3}
Add 1 to 60.
x=\frac{1±\sqrt{61}}{2\times 3}
The opposite of -1 is 1.
x=\frac{1±\sqrt{61}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{61}+1}{6}
Now solve the equation x=\frac{1±\sqrt{61}}{6} when ± is plus. Add 1 to \sqrt{61}.
x=\frac{1-\sqrt{61}}{6}
Now solve the equation x=\frac{1±\sqrt{61}}{6} when ± is minus. Subtract \sqrt{61} from 1.
x=\frac{\sqrt{61}+1}{6} x=\frac{1-\sqrt{61}}{6}
The equation is now solved.
3x^{2}-x=5
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}-x}{3}=\frac{5}{3}
Divide both sides by 3.
x^{2}-\frac{1}{3}x=\frac{5}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{5}{3}+\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^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{5}{3}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{61}{36}
Add \frac{5}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{6}\right)^{2}=\frac{61}{36}
Factor x^{2}-\frac{1}{3}x+\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-\frac{1}{6}\right)^{2}}=\sqrt{\frac{61}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{\sqrt{61}}{6} x-\frac{1}{6}=-\frac{\sqrt{61}}{6}
Simplify.
x=\frac{\sqrt{61}+1}{6} x=\frac{1-\sqrt{61}}{6}
Add \frac{1}{6} to both sides of the equation.