# Find the roots of the quadratic polynomials:f(x)=x^2-2x-1

Find the roots of the quadratic polynomials:
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}$$

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We know that, for an standard equation $$\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{2}}+{b}{x}+{c}$$, roots are calculated by using quadratic formula as $$\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}$$
We have the given equation as
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}$$
On comparing the equation $$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}$$ with standard equation $$\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{2}}+{b}{x}+{c}$$, we get the result as $$\displaystyle{a}={1},{b}=-{2},{\quad\text{and}\quad}{c}=-{1}$$
Therefore, by using equation, we get the zeros as
$$\displaystyle{x}={\frac{{-{\left(-{2}\right)}\pm\sqrt{{{\left(-{2}\right)}^{{2}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{\left({2}\right)}{\left({1}\right)}}}}$$
$$\displaystyle{x}={\frac{{{2}\pm\sqrt{{{4}+{4}}}}}{{{2}}}}$$
$$\displaystyle{x}={\frac{{{2}\pm{2}\sqrt{{{2}}}}}{{{2}}}}$$
$$\displaystyle{x}={1}\pm\sqrt{{{2}}}$$
Hence, zeros of the quadratic $$\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}$$ are $$\displaystyle{x}={1}+\sqrt{{{2}}}$$  and  $$\displaystyle{x}={1}-\sqrt{{{2}}}$$