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Solve for x (complex solution)
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-x^{2}-x-6=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(-1\right)±\sqrt{1-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+4\left(-6\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-1\right)±\sqrt{1-24}}{2\left(-1\right)}
Multiply 4 times -6.
x=\frac{-\left(-1\right)±\sqrt{-23}}{2\left(-1\right)}
Add 1 to -24.
x=\frac{-\left(-1\right)±\sqrt{23}i}{2\left(-1\right)}
Take the square root of -23.
x=\frac{1±\sqrt{23}i}{2\left(-1\right)}
The opposite of -1 is 1.
x=\frac{1±\sqrt{23}i}{-2}
Multiply 2 times -1.
x=\frac{1+\sqrt{23}i}{-2}
Now solve the equation x=\frac{1±\sqrt{23}i}{-2} when ± is plus. Add 1 to i\sqrt{23}.
x=\frac{-\sqrt{23}i-1}{2}
Divide 1+i\sqrt{23} by -2.
x=\frac{-\sqrt{23}i+1}{-2}
Now solve the equation x=\frac{1±\sqrt{23}i}{-2} when ± is minus. Subtract i\sqrt{23} from 1.
x=\frac{-1+\sqrt{23}i}{2}
Divide 1-i\sqrt{23} by -2.
x=\frac{-\sqrt{23}i-1}{2} x=\frac{-1+\sqrt{23}i}{2}
The equation is now solved.
-x^{2}-x-6=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}-x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
-x^{2}-x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
-x^{2}-x=6
Subtract -6 from 0.
\frac{-x^{2}-x}{-1}=\frac{6}{-1}
Divide both sides by -1.
x^{2}+\left(-\frac{1}{-1}\right)x=\frac{6}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+x=\frac{6}{-1}
Divide -1 by -1.
x^{2}+x=-6
Divide 6 by -1.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-6+\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.
x^{2}+x+\frac{1}{4}=-6+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=-\frac{23}{4}
Add -6 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=-\frac{23}{4}
Factor x^{2}+x+\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(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{23}i}{2} x+\frac{1}{2}=-\frac{\sqrt{23}i}{2}
Simplify.
x=\frac{-1+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i-1}{2}
Subtract \frac{1}{2} from both sides of the equation.