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x^{2}-2x-\frac{5}{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.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-\frac{5}{3}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -\frac{5}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\left(-\frac{5}{3}\right)}}{2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+\frac{20}{3}}}{2}
Multiply -4 times -\frac{5}{3}.
x=\frac{-\left(-2\right)±\sqrt{\frac{32}{3}}}{2}
Add 4 to \frac{20}{3}.
x=\frac{-\left(-2\right)±\frac{4\sqrt{6}}{3}}{2}
Take the square root of \frac{32}{3}.
x=\frac{2±\frac{4\sqrt{6}}{3}}{2}
The opposite of -2 is 2.
x=\frac{\frac{4\sqrt{6}}{3}+2}{2}
Now solve the equation x=\frac{2±\frac{4\sqrt{6}}{3}}{2} when ± is plus. Add 2 to \frac{4\sqrt{6}}{3}.
x=\frac{2\sqrt{6}}{3}+1
Divide 2+\frac{4\sqrt{6}}{3} by 2.
x=\frac{-\frac{4\sqrt{6}}{3}+2}{2}
Now solve the equation x=\frac{2±\frac{4\sqrt{6}}{3}}{2} when ± is minus. Subtract \frac{4\sqrt{6}}{3} from 2.
x=-\frac{2\sqrt{6}}{3}+1
Divide 2-\frac{4\sqrt{6}}{3} by 2.
x=\frac{2\sqrt{6}}{3}+1 x=-\frac{2\sqrt{6}}{3}+1
The equation is now solved.
x^{2}-2x-\frac{5}{3}=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.
x^{2}-2x-\frac{5}{3}-\left(-\frac{5}{3}\right)=-\left(-\frac{5}{3}\right)
Add \frac{5}{3} to both sides of the equation.
x^{2}-2x=-\left(-\frac{5}{3}\right)
Subtracting -\frac{5}{3} from itself leaves 0.
x^{2}-2x=\frac{5}{3}
Subtract -\frac{5}{3} from 0.
x^{2}-2x+1=\frac{5}{3}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=\frac{8}{3}
Add \frac{5}{3} to 1.
\left(x-1\right)^{2}=\frac{8}{3}
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{\frac{8}{3}}
Take the square root of both sides of the equation.
x-1=\frac{2\sqrt{6}}{3} x-1=-\frac{2\sqrt{6}}{3}
Simplify.
x=\frac{2\sqrt{6}}{3}+1 x=-\frac{2\sqrt{6}}{3}+1
Add 1 to both sides of the equation.