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Solve for x (complex solution)
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2x+5+x^{2}=0
Add x^{2} to both sides.
x^{2}+2x+5=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{-2±\sqrt{2^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 5}}{2}
Square 2.
x=\frac{-2±\sqrt{4-20}}{2}
Multiply -4 times 5.
x=\frac{-2±\sqrt{-16}}{2}
Add 4 to -20.
x=\frac{-2±4i}{2}
Take the square root of -16.
x=\frac{-2+4i}{2}
Now solve the equation x=\frac{-2±4i}{2} when ± is plus. Add -2 to 4i.
x=-1+2i
Divide -2+4i by 2.
x=\frac{-2-4i}{2}
Now solve the equation x=\frac{-2±4i}{2} when ± is minus. Subtract 4i from -2.
x=-1-2i
Divide -2-4i by 2.
x=-1+2i x=-1-2i
The equation is now solved.
2x+5+x^{2}=0
Add x^{2} to both sides.
2x+x^{2}=-5
Subtract 5 from both sides. Anything subtracted from zero gives its negation.
x^{2}+2x=-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.
x^{2}+2x+1^{2}=-5+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.
x^{2}+2x+1=-5+1
Square 1.
x^{2}+2x+1=-4
Add -5 to 1.
\left(x+1\right)^{2}=-4
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{-4}
Take the square root of both sides of the equation.
x+1=2i x+1=-2i
Simplify.
x=-1+2i x=-1-2i
Subtract 1 from both sides of the equation.