Skip to main content
Solve for n
Tick mark Image

Similar Problems from Web Search

Share

2n^{2}-\frac{8}{3}n+\frac{7}{9}=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.
n=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\left(-\frac{8}{3}\right)^{2}-4\times 2\times \frac{7}{9}}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -\frac{8}{3} for b, and \frac{7}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-4\times 2\times \frac{7}{9}}}{2\times 2}
Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.
n=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-8\times \frac{7}{9}}}{2\times 2}
Multiply -4 times 2.
n=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64-56}{9}}}{2\times 2}
Multiply -8 times \frac{7}{9}.
n=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{8}{9}}}{2\times 2}
Add \frac{64}{9} to -\frac{56}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
n=\frac{-\left(-\frac{8}{3}\right)±\frac{2\sqrt{2}}{3}}{2\times 2}
Take the square root of \frac{8}{9}.
n=\frac{\frac{8}{3}±\frac{2\sqrt{2}}{3}}{2\times 2}
The opposite of -\frac{8}{3} is \frac{8}{3}.
n=\frac{\frac{8}{3}±\frac{2\sqrt{2}}{3}}{4}
Multiply 2 times 2.
n=\frac{2\sqrt{2}+8}{3\times 4}
Now solve the equation n=\frac{\frac{8}{3}±\frac{2\sqrt{2}}{3}}{4} when ± is plus. Add \frac{8}{3} to \frac{2\sqrt{2}}{3}.
n=\frac{\sqrt{2}}{6}+\frac{2}{3}
Divide \frac{8+2\sqrt{2}}{3} by 4.
n=\frac{8-2\sqrt{2}}{3\times 4}
Now solve the equation n=\frac{\frac{8}{3}±\frac{2\sqrt{2}}{3}}{4} when ± is minus. Subtract \frac{2\sqrt{2}}{3} from \frac{8}{3}.
n=-\frac{\sqrt{2}}{6}+\frac{2}{3}
Divide \frac{8-2\sqrt{2}}{3} by 4.
n=\frac{\sqrt{2}}{6}+\frac{2}{3} n=-\frac{\sqrt{2}}{6}+\frac{2}{3}
The equation is now solved.
2n^{2}-\frac{8}{3}n+\frac{7}{9}=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.
2n^{2}-\frac{8}{3}n+\frac{7}{9}-\frac{7}{9}=-\frac{7}{9}
Subtract \frac{7}{9} from both sides of the equation.
2n^{2}-\frac{8}{3}n=-\frac{7}{9}
Subtracting \frac{7}{9} from itself leaves 0.
\frac{2n^{2}-\frac{8}{3}n}{2}=-\frac{\frac{7}{9}}{2}
Divide both sides by 2.
n^{2}+\left(-\frac{\frac{8}{3}}{2}\right)n=-\frac{\frac{7}{9}}{2}
Dividing by 2 undoes the multiplication by 2.
n^{2}-\frac{4}{3}n=-\frac{\frac{7}{9}}{2}
Divide -\frac{8}{3} by 2.
n^{2}-\frac{4}{3}n=-\frac{7}{18}
Divide -\frac{7}{9} by 2.
n^{2}-\frac{4}{3}n+\left(-\frac{2}{3}\right)^{2}=-\frac{7}{18}+\left(-\frac{2}{3}\right)^{2}
Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}-\frac{4}{3}n+\frac{4}{9}=-\frac{7}{18}+\frac{4}{9}
Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.
n^{2}-\frac{4}{3}n+\frac{4}{9}=\frac{1}{18}
Add -\frac{7}{18} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(n-\frac{2}{3}\right)^{2}=\frac{1}{18}
Factor n^{2}-\frac{4}{3}n+\frac{4}{9}. 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(n-\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{18}}
Take the square root of both sides of the equation.
n-\frac{2}{3}=\frac{\sqrt{2}}{6} n-\frac{2}{3}=-\frac{\sqrt{2}}{6}
Simplify.
n=\frac{\sqrt{2}}{6}+\frac{2}{3} n=-\frac{\sqrt{2}}{6}+\frac{2}{3}
Add \frac{2}{3} to both sides of the equation.