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