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\frac{1}{2}x^{2}-x-\frac{3}{2}=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(-1\right)±\sqrt{1-4\times \frac{1}{2}\left(-\frac{3}{2}\right)}}{2\times \frac{1}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -1 for b, and -\frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-2\left(-\frac{3}{2}\right)}}{2\times \frac{1}{2}}
Multiply -4 times \frac{1}{2}.
x=\frac{-\left(-1\right)±\sqrt{1+3}}{2\times \frac{1}{2}}
Multiply -2 times -\frac{3}{2}.
x=\frac{-\left(-1\right)±\sqrt{4}}{2\times \frac{1}{2}}
Add 1 to 3.
x=\frac{-\left(-1\right)±2}{2\times \frac{1}{2}}
Take the square root of 4.
x=\frac{1±2}{2\times \frac{1}{2}}
The opposite of -1 is 1.
x=\frac{1±2}{1}
Multiply 2 times \frac{1}{2}.
x=\frac{3}{1}
Now solve the equation x=\frac{1±2}{1} when ± is plus. Add 1 to 2.
x=3
Divide 3 by 1.
x=-\frac{1}{1}
Now solve the equation x=\frac{1±2}{1} when ± is minus. Subtract 2 from 1.
x=-1
Divide -1 by 1.
x=3 x=-1
The equation is now solved.
\frac{1}{2}x^{2}-x-\frac{3}{2}=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}-x-\frac{3}{2}-\left(-\frac{3}{2}\right)=-\left(-\frac{3}{2}\right)
Add \frac{3}{2} to both sides of the equation.
\frac{1}{2}x^{2}-x=-\left(-\frac{3}{2}\right)
Subtracting -\frac{3}{2} from itself leaves 0.
\frac{1}{2}x^{2}-x=\frac{3}{2}
Subtract -\frac{3}{2} from 0.
\frac{\frac{1}{2}x^{2}-x}{\frac{1}{2}}=\frac{\frac{3}{2}}{\frac{1}{2}}
Multiply both sides by 2.
x^{2}+\left(-\frac{1}{\frac{1}{2}}\right)x=\frac{\frac{3}{2}}{\frac{1}{2}}
Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.
x^{2}-2x=\frac{\frac{3}{2}}{\frac{1}{2}}
Divide -1 by \frac{1}{2} by multiplying -1 by the reciprocal of \frac{1}{2}.
x^{2}-2x=3
Divide \frac{3}{2} by \frac{1}{2} by multiplying \frac{3}{2} by the reciprocal of \frac{1}{2}.
x^{2}-2x+1=3+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=4
Add 3 to 1.
\left(x-1\right)^{2}=4
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x-1=2 x-1=-2
Simplify.
x=3 x=-1
Add 1 to both sides of the equation.