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x^{2}-x=3
Subtract x from both sides.
x^{2}-x-3=0
Subtract 3 from both sides.
x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+12}}{2}
Multiply -4 times -3.
x=\frac{-\left(-1\right)±\sqrt{13}}{2}
Add 1 to 12.
x=\frac{1±\sqrt{13}}{2}
The opposite of -1 is 1.
x=\frac{\sqrt{13}+1}{2}
Now solve the equation x=\frac{1±\sqrt{13}}{2} when ± is plus. Add 1 to \sqrt{13}.
x=\frac{1-\sqrt{13}}{2}
Now solve the equation x=\frac{1±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from 1.
x=\frac{\sqrt{13}+1}{2} x=\frac{1-\sqrt{13}}{2}
The equation is now solved.
x^{2}-x=3
Subtract x from both sides.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=3+\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}=3+\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{13}{4}
Add 3 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=\frac{13}{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{13}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{13}}{2} x-\frac{1}{2}=-\frac{\sqrt{13}}{2}
Simplify.
x=\frac{\sqrt{13}+1}{2} x=\frac{1-\sqrt{13}}{2}
Add \frac{1}{2} to both sides of the equation.