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2x^{2}-2x=1
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.
2x^{2}-2x-1=1-1
Subtract 1 from both sides of the equation.
2x^{2}-2x-1=0
Subtracting 1 from itself leaves 0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-1\right)}}{2\times 2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-8\left(-1\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-2\right)±\sqrt{4+8}}{2\times 2}
Multiply -8 times -1.
x=\frac{-\left(-2\right)±\sqrt{12}}{2\times 2}
Add 4 to 8.
x=\frac{-\left(-2\right)±2\sqrt{3}}{2\times 2}
Take the square root of 12.
x=\frac{2±2\sqrt{3}}{2\times 2}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{3}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{3}+2}{4}
Now solve the equation x=\frac{2±2\sqrt{3}}{4} when ± is plus. Add 2 to 2\sqrt{3}.
x=\frac{\sqrt{3}+1}{2}
Divide 2+2\sqrt{3} by 4.
x=\frac{2-2\sqrt{3}}{4}
Now solve the equation x=\frac{2±2\sqrt{3}}{4} when ± is minus. Subtract 2\sqrt{3} from 2.
x=\frac{1-\sqrt{3}}{2}
Divide 2-2\sqrt{3} by 4.
x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}
The equation is now solved.
2x^{2}-2x=1
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{2x^{2}-2x}{2}=\frac{1}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{2}{2}\right)x=\frac{1}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-x=\frac{1}{2}
Divide -2 by 2.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{1}{2}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=\frac{1}{2}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{3}{4}
Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}
Factor x^{2}-x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{3}}{2} x-\frac{1}{2}=-\frac{\sqrt{3}}{2}
Simplify.
x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}
Add \frac{1}{2} to both sides of the equation.