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Solve for x (complex solution)
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0.6x^{2}-0.3x+0.3=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(-0.3\right)±\sqrt{\left(-0.3\right)^{2}-4\times 0.6\times 0.3}}{2\times 0.6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.6 for a, -0.3 for b, and 0.3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-0.3\right)±\sqrt{0.09-4\times 0.6\times 0.3}}{2\times 0.6}
Square -0.3 by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-0.3\right)±\sqrt{0.09-2.4\times 0.3}}{2\times 0.6}
Multiply -4 times 0.6.
x=\frac{-\left(-0.3\right)±\sqrt{0.09-0.72}}{2\times 0.6}
Multiply -2.4 times 0.3 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-0.3\right)±\sqrt{-0.63}}{2\times 0.6}
Add 0.09 to -0.72 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-0.3\right)±\frac{3\sqrt{7}i}{10}}{2\times 0.6}
Take the square root of -0.63.
x=\frac{0.3±\frac{3\sqrt{7}i}{10}}{2\times 0.6}
The opposite of -0.3 is 0.3.
x=\frac{0.3±\frac{3\sqrt{7}i}{10}}{1.2}
Multiply 2 times 0.6.
x=\frac{3+3\sqrt{7}i}{1.2\times 10}
Now solve the equation x=\frac{0.3±\frac{3\sqrt{7}i}{10}}{1.2} when ± is plus. Add 0.3 to \frac{3i\sqrt{7}}{10}.
x=\frac{1+\sqrt{7}i}{4}
Divide \frac{3+3i\sqrt{7}}{10} by 1.2 by multiplying \frac{3+3i\sqrt{7}}{10} by the reciprocal of 1.2.
x=\frac{-3\sqrt{7}i+3}{1.2\times 10}
Now solve the equation x=\frac{0.3±\frac{3\sqrt{7}i}{10}}{1.2} when ± is minus. Subtract \frac{3i\sqrt{7}}{10} from 0.3.
x=\frac{-\sqrt{7}i+1}{4}
Divide \frac{3-3i\sqrt{7}}{10} by 1.2 by multiplying \frac{3-3i\sqrt{7}}{10} by the reciprocal of 1.2.
x=\frac{1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+1}{4}
The equation is now solved.
0.6x^{2}-0.3x+0.3=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.
0.6x^{2}-0.3x+0.3-0.3=-0.3
Subtract 0.3 from both sides of the equation.
0.6x^{2}-0.3x=-0.3
Subtracting 0.3 from itself leaves 0.
\frac{0.6x^{2}-0.3x}{0.6}=-\frac{0.3}{0.6}
Divide both sides of the equation by 0.6, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\left(-\frac{0.3}{0.6}\right)x=-\frac{0.3}{0.6}
Dividing by 0.6 undoes the multiplication by 0.6.
x^{2}-0.5x=-\frac{0.3}{0.6}
Divide -0.3 by 0.6 by multiplying -0.3 by the reciprocal of 0.6.
x^{2}-0.5x=-0.5
Divide -0.3 by 0.6 by multiplying -0.3 by the reciprocal of 0.6.
x^{2}-0.5x+\left(-0.25\right)^{2}=-0.5+\left(-0.25\right)^{2}
Divide -0.5, the coefficient of the x term, by 2 to get -0.25. Then add the square of -0.25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-0.5x+0.0625=-0.5+0.0625
Square -0.25 by squaring both the numerator and the denominator of the fraction.
x^{2}-0.5x+0.0625=-0.4375
Add -0.5 to 0.0625 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-0.25\right)^{2}=-0.4375
Factor x^{2}-0.5x+0.0625. 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-0.25\right)^{2}}=\sqrt{-0.4375}
Take the square root of both sides of the equation.
x-0.25=\frac{\sqrt{7}i}{4} x-0.25=-\frac{\sqrt{7}i}{4}
Simplify.
x=\frac{1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+1}{4}
Add 0.25 to both sides of the equation.