Skip to main content
Solve for x (complex solution)
Tick mark Image
Graph

Similar Problems from Web Search

Share

6x^{2}-\frac{16}{3}x+\frac{4}{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(-\frac{16}{3}\right)±\sqrt{\left(-\frac{16}{3}\right)^{2}-4\times 6\times \frac{4}{3}}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -\frac{16}{3} for b, and \frac{4}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256}{9}-4\times 6\times \frac{4}{3}}}{2\times 6}
Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256}{9}-24\times \frac{4}{3}}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-\frac{16}{3}\right)±\sqrt{\frac{256}{9}-32}}{2\times 6}
Multiply -24 times \frac{4}{3}.
x=\frac{-\left(-\frac{16}{3}\right)±\sqrt{-\frac{32}{9}}}{2\times 6}
Add \frac{256}{9} to -32.
x=\frac{-\left(-\frac{16}{3}\right)±\frac{4\sqrt{2}i}{3}}{2\times 6}
Take the square root of -\frac{32}{9}.
x=\frac{\frac{16}{3}±\frac{4\sqrt{2}i}{3}}{2\times 6}
The opposite of -\frac{16}{3} is \frac{16}{3}.
x=\frac{\frac{16}{3}±\frac{4\sqrt{2}i}{3}}{12}
Multiply 2 times 6.
x=\frac{16+4\sqrt{2}i}{3\times 12}
Now solve the equation x=\frac{\frac{16}{3}±\frac{4\sqrt{2}i}{3}}{12} when ± is plus. Add \frac{16}{3} to \frac{4i\sqrt{2}}{3}.
x=\frac{4+\sqrt{2}i}{9}
Divide \frac{16+4i\sqrt{2}}{3} by 12.
x=\frac{-4\sqrt{2}i+16}{3\times 12}
Now solve the equation x=\frac{\frac{16}{3}±\frac{4\sqrt{2}i}{3}}{12} when ± is minus. Subtract \frac{4i\sqrt{2}}{3} from \frac{16}{3}.
x=\frac{-\sqrt{2}i+4}{9}
Divide \frac{16-4i\sqrt{2}}{3} by 12.
x=\frac{4+\sqrt{2}i}{9} x=\frac{-\sqrt{2}i+4}{9}
The equation is now solved.
6x^{2}-\frac{16}{3}x+\frac{4}{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.
6x^{2}-\frac{16}{3}x+\frac{4}{3}-\frac{4}{3}=-\frac{4}{3}
Subtract \frac{4}{3} from both sides of the equation.
6x^{2}-\frac{16}{3}x=-\frac{4}{3}
Subtracting \frac{4}{3} from itself leaves 0.
\frac{6x^{2}-\frac{16}{3}x}{6}=-\frac{\frac{4}{3}}{6}
Divide both sides by 6.
x^{2}+\left(-\frac{\frac{16}{3}}{6}\right)x=-\frac{\frac{4}{3}}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{8}{9}x=-\frac{\frac{4}{3}}{6}
Divide -\frac{16}{3} by 6.
x^{2}-\frac{8}{9}x=-\frac{2}{9}
Divide -\frac{4}{3} by 6.
x^{2}-\frac{8}{9}x+\left(-\frac{4}{9}\right)^{2}=-\frac{2}{9}+\left(-\frac{4}{9}\right)^{2}
Divide -\frac{8}{9}, the coefficient of the x term, by 2 to get -\frac{4}{9}. Then add the square of -\frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{8}{9}x+\frac{16}{81}=-\frac{2}{9}+\frac{16}{81}
Square -\frac{4}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{8}{9}x+\frac{16}{81}=-\frac{2}{81}
Add -\frac{2}{9} to \frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{4}{9}\right)^{2}=-\frac{2}{81}
Factor x^{2}-\frac{8}{9}x+\frac{16}{81}. 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{4}{9}\right)^{2}}=\sqrt{-\frac{2}{81}}
Take the square root of both sides of the equation.
x-\frac{4}{9}=\frac{\sqrt{2}i}{9} x-\frac{4}{9}=-\frac{\sqrt{2}i}{9}
Simplify.
x=\frac{4+\sqrt{2}i}{9} x=\frac{-\sqrt{2}i+4}{9}
Add \frac{4}{9} to both sides of the equation.