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Solve for x (complex solution)
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2^{2}x^{2}+x+1=0
Expand \left(2x\right)^{2}.
4x^{2}+x+1=0
Calculate 2 to the power of 2 and get 4.
x=\frac{-1±\sqrt{1^{2}-4\times 4}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\times 4}}{2\times 4}
Square 1.
x=\frac{-1±\sqrt{1-16}}{2\times 4}
Multiply -4 times 4.
x=\frac{-1±\sqrt{-15}}{2\times 4}
Add 1 to -16.
x=\frac{-1±\sqrt{15}i}{2\times 4}
Take the square root of -15.
x=\frac{-1±\sqrt{15}i}{8}
Multiply 2 times 4.
x=\frac{-1+\sqrt{15}i}{8}
Now solve the equation x=\frac{-1±\sqrt{15}i}{8} when ± is plus. Add -1 to i\sqrt{15}.
x=\frac{-\sqrt{15}i-1}{8}
Now solve the equation x=\frac{-1±\sqrt{15}i}{8} when ± is minus. Subtract i\sqrt{15} from -1.
x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}
The equation is now solved.
2^{2}x^{2}+x+1=0
Expand \left(2x\right)^{2}.
4x^{2}+x+1=0
Calculate 2 to the power of 2 and get 4.
4x^{2}+x=-1
Subtract 1 from both sides. Anything subtracted from zero gives its negation.
\frac{4x^{2}+x}{4}=-\frac{1}{4}
Divide both sides by 4.
x^{2}+\frac{1}{4}x=-\frac{1}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=-\frac{1}{4}+\left(\frac{1}{8}\right)^{2}
Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{1}{4}+\frac{1}{64}
Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{15}{64}
Add -\frac{1}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{8}\right)^{2}=-\frac{15}{64}
Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. 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}{8}\right)^{2}}=\sqrt{-\frac{15}{64}}
Take the square root of both sides of the equation.
x+\frac{1}{8}=\frac{\sqrt{15}i}{8} x+\frac{1}{8}=-\frac{\sqrt{15}i}{8}
Simplify.
x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}
Subtract \frac{1}{8} from both sides of the equation.