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2x^{2}-\frac{3}{2}x+\frac{7}{10}=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{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\times 2\times \frac{7}{10}}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -\frac{3}{2} for b, and \frac{7}{10} 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 2\times \frac{7}{10}}}{2\times 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}-8\times \frac{7}{10}}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-\frac{28}{5}}}{2\times 2}
Multiply -8 times \frac{7}{10}.
x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{-\frac{67}{20}}}{2\times 2}
Add \frac{9}{4} to -\frac{28}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-\frac{3}{2}\right)±\frac{\sqrt{335}i}{10}}{2\times 2}
Take the square root of -\frac{67}{20}.
x=\frac{\frac{3}{2}±\frac{\sqrt{335}i}{10}}{2\times 2}
The opposite of -\frac{3}{2} is \frac{3}{2}.
x=\frac{\frac{3}{2}±\frac{\sqrt{335}i}{10}}{4}
Multiply 2 times 2.
x=\frac{\frac{\sqrt{335}i}{10}+\frac{3}{2}}{4}
Now solve the equation x=\frac{\frac{3}{2}±\frac{\sqrt{335}i}{10}}{4} when ± is plus. Add \frac{3}{2} to \frac{i\sqrt{335}}{10}.
x=\frac{\sqrt{335}i}{40}+\frac{3}{8}
Divide \frac{3}{2}+\frac{i\sqrt{335}}{10} by 4.
x=\frac{-\frac{\sqrt{335}i}{10}+\frac{3}{2}}{4}
Now solve the equation x=\frac{\frac{3}{2}±\frac{\sqrt{335}i}{10}}{4} when ± is minus. Subtract \frac{i\sqrt{335}}{10} from \frac{3}{2}.
x=-\frac{\sqrt{335}i}{40}+\frac{3}{8}
Divide \frac{3}{2}-\frac{i\sqrt{335}}{10} by 4.
x=\frac{\sqrt{335}i}{40}+\frac{3}{8} x=-\frac{\sqrt{335}i}{40}+\frac{3}{8}
The equation is now solved.
2x^{2}-\frac{3}{2}x+\frac{7}{10}=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.
2x^{2}-\frac{3}{2}x+\frac{7}{10}-\frac{7}{10}=-\frac{7}{10}
Subtract \frac{7}{10} from both sides of the equation.
2x^{2}-\frac{3}{2}x=-\frac{7}{10}
Subtracting \frac{7}{10} from itself leaves 0.
\frac{2x^{2}-\frac{3}{2}x}{2}=-\frac{\frac{7}{10}}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{\frac{3}{2}}{2}\right)x=-\frac{\frac{7}{10}}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{3}{4}x=-\frac{\frac{7}{10}}{2}
Divide -\frac{3}{2} by 2.
x^{2}-\frac{3}{4}x=-\frac{7}{20}
Divide -\frac{7}{10} by 2.
x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-\frac{7}{20}+\left(-\frac{3}{8}\right)^{2}
Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{7}{20}+\frac{9}{64}
Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{67}{320}
Add -\frac{7}{20} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{8}\right)^{2}=-\frac{67}{320}
Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. 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}{8}\right)^{2}}=\sqrt{-\frac{67}{320}}
Take the square root of both sides of the equation.
x-\frac{3}{8}=\frac{\sqrt{335}i}{40} x-\frac{3}{8}=-\frac{\sqrt{335}i}{40}
Simplify.
x=\frac{\sqrt{335}i}{40}+\frac{3}{8} x=-\frac{\sqrt{335}i}{40}+\frac{3}{8}
Add \frac{3}{8} to both sides of the equation.