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