Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

\frac{1}{2}x^{2}-\frac{5}{3}x-8=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{5}{3}\right)±\sqrt{\left(-\frac{5}{3}\right)^{2}-4\times \frac{1}{2}\left(-8\right)}}{2\times \frac{1}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{5}{3} for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-4\times \frac{1}{2}\left(-8\right)}}{2\times \frac{1}{2}}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-2\left(-8\right)}}{2\times \frac{1}{2}}
Multiply -4 times \frac{1}{2}.
x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}+16}}{2\times \frac{1}{2}}
Multiply -2 times -8.
x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{169}{9}}}{2\times \frac{1}{2}}
Add \frac{25}{9} to 16.
x=\frac{-\left(-\frac{5}{3}\right)±\frac{13}{3}}{2\times \frac{1}{2}}
Take the square root of \frac{169}{9}.
x=\frac{\frac{5}{3}±\frac{13}{3}}{2\times \frac{1}{2}}
The opposite of -\frac{5}{3} is \frac{5}{3}.
x=\frac{\frac{5}{3}±\frac{13}{3}}{1}
Multiply 2 times \frac{1}{2}.
x=\frac{6}{1}
Now solve the equation x=\frac{\frac{5}{3}±\frac{13}{3}}{1} when ± is plus. Add \frac{5}{3} to \frac{13}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=6
Divide 6 by 1.
x=-\frac{\frac{8}{3}}{1}
Now solve the equation x=\frac{\frac{5}{3}±\frac{13}{3}}{1} when ± is minus. Subtract \frac{13}{3} from \frac{5}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
x=-\frac{8}{3}
Divide -\frac{8}{3} by 1.
x=6 x=-\frac{8}{3}
The equation is now solved.
\frac{1}{2}x^{2}-\frac{5}{3}x-8=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.
\frac{1}{2}x^{2}-\frac{5}{3}x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
\frac{1}{2}x^{2}-\frac{5}{3}x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
\frac{1}{2}x^{2}-\frac{5}{3}x=8
Subtract -8 from 0.
\frac{\frac{1}{2}x^{2}-\frac{5}{3}x}{\frac{1}{2}}=\frac{8}{\frac{1}{2}}
Multiply both sides by 2.
x^{2}+\left(-\frac{\frac{5}{3}}{\frac{1}{2}}\right)x=\frac{8}{\frac{1}{2}}
Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.
x^{2}-\frac{10}{3}x=\frac{8}{\frac{1}{2}}
Divide -\frac{5}{3} by \frac{1}{2} by multiplying -\frac{5}{3} by the reciprocal of \frac{1}{2}.
x^{2}-\frac{10}{3}x=16
Divide 8 by \frac{1}{2} by multiplying 8 by the reciprocal of \frac{1}{2}.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=16+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=16+\frac{25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{169}{9}
Add 16 to \frac{25}{9}.
\left(x-\frac{5}{3}\right)^{2}=\frac{169}{9}
Factor x^{2}-\frac{10}{3}x+\frac{25}{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(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{169}{9}}
Take the square root of both sides of the equation.
x-\frac{5}{3}=\frac{13}{3} x-\frac{5}{3}=-\frac{13}{3}
Simplify.
x=6 x=-\frac{8}{3}
Add \frac{5}{3} to both sides of the equation.