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s^{2}+2s+2=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.
s=\frac{-2±\sqrt{2^{2}-4\times 2}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-2±\sqrt{4-4\times 2}}{2}
Square 2.
s=\frac{-2±\sqrt{4-8}}{2}
Multiply -4 times 2.
s=\frac{-2±\sqrt{-4}}{2}
Add 4 to -8.
s=\frac{-2±2i}{2}
Take the square root of -4.
s=\frac{-2+2i}{2}
Now solve the equation s=\frac{-2±2i}{2} when ± is plus. Add -2 to 2i.
s=-1+i
Divide -2+2i by 2.
s=\frac{-2-2i}{2}
Now solve the equation s=\frac{-2±2i}{2} when ± is minus. Subtract 2i from -2.
s=-1-i
Divide -2-2i by 2.
s=-1+i s=-1-i
The equation is now solved.
s^{2}+2s+2=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.
s^{2}+2s+2-2=-2
Subtract 2 from both sides of the equation.
s^{2}+2s=-2
Subtracting 2 from itself leaves 0.
s^{2}+2s+1^{2}=-2+1^{2}
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.
s^{2}+2s+1=-2+1
Square 1.
s^{2}+2s+1=-1
Add -2 to 1.
\left(s+1\right)^{2}=-1
Factor s^{2}+2s+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(s+1\right)^{2}}=\sqrt{-1}
Take the square root of both sides of the equation.
s+1=i s+1=-i
Simplify.
s=-1+i s=-1-i
Subtract 1 from both sides of the equation.