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Solve for x (complex solution)
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-x^{2}+x+1=5
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^{2}+x+1-5=5-5
Subtract 5 from both sides of the equation.
-x^{2}+x+1-5=0
Subtracting 5 from itself leaves 0.
-x^{2}+x-4=0
Subtract 5 from 1.
x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}
Square 1.
x=\frac{-1±\sqrt{1+4\left(-4\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-1±\sqrt{1-16}}{2\left(-1\right)}
Multiply 4 times -4.
x=\frac{-1±\sqrt{-15}}{2\left(-1\right)}
Add 1 to -16.
x=\frac{-1±\sqrt{15}i}{2\left(-1\right)}
Take the square root of -15.
x=\frac{-1±\sqrt{15}i}{-2}
Multiply 2 times -1.
x=\frac{-1+\sqrt{15}i}{-2}
Now solve the equation x=\frac{-1±\sqrt{15}i}{-2} when ± is plus. Add -1 to i\sqrt{15}.
x=\frac{-\sqrt{15}i+1}{2}
Divide -1+i\sqrt{15} by -2.
x=\frac{-\sqrt{15}i-1}{-2}
Now solve the equation x=\frac{-1±\sqrt{15}i}{-2} when ± is minus. Subtract i\sqrt{15} from -1.
x=\frac{1+\sqrt{15}i}{2}
Divide -1-i\sqrt{15} by -2.
x=\frac{-\sqrt{15}i+1}{2} x=\frac{1+\sqrt{15}i}{2}
The equation is now solved.
-x^{2}+x+1=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}+x+1-1=5-1
Subtract 1 from both sides of the equation.
-x^{2}+x=5-1
Subtracting 1 from itself leaves 0.
-x^{2}+x=4
Subtract 1 from 5.
\frac{-x^{2}+x}{-1}=\frac{4}{-1}
Divide both sides by -1.
x^{2}+\frac{1}{-1}x=\frac{4}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}-x=\frac{4}{-1}
Divide 1 by -1.
x^{2}-x=-4
Divide 4 by -1.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-4+\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}=-4+\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{15}{4}
Add -4 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=-\frac{15}{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{15}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{15}i}{2} x-\frac{1}{2}=-\frac{\sqrt{15}i}{2}
Simplify.
x=\frac{1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+1}{2}
Add \frac{1}{2} to both sides of the equation.