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Solve for x (complex solution)
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x^{2}+6x+13=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{-6±\sqrt{6^{2}-4\times 13}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 13}}{2}
Square 6.
x=\frac{-6±\sqrt{36-52}}{2}
Multiply -4 times 13.
x=\frac{-6±\sqrt{-16}}{2}
Add 36 to -52.
x=\frac{-6±4i}{2}
Take the square root of -16.
x=\frac{-6+4i}{2}
Now solve the equation x=\frac{-6±4i}{2} when ± is plus. Add -6 to 4i.
x=-3+2i
Divide -6+4i by 2.
x=\frac{-6-4i}{2}
Now solve the equation x=\frac{-6±4i}{2} when ± is minus. Subtract 4i from -6.
x=-3-2i
Divide -6-4i by 2.
x=-3+2i x=-3-2i
The equation is now solved.
x^{2}+6x+13=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}+6x+13-13=-13
Subtract 13 from both sides of the equation.
x^{2}+6x=-13
Subtracting 13 from itself leaves 0.
x^{2}+6x+3^{2}=-13+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+6x+9=-13+9
Square 3.
x^{2}+6x+9=-4
Add -13 to 9.
\left(x+3\right)^{2}=-4
Factor x^{2}+6x+9. 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+3\right)^{2}}=\sqrt{-4}
Take the square root of both sides of the equation.
x+3=2i x+3=-2i
Simplify.
x=-3+2i x=-3-2i
Subtract 3 from both sides of the equation.