# Find the roots of the quadratic polynomials:6x^2+5x-1 , 4x^2+4x-4

Find the roots of the quadratic polynomials:
$$\displaystyle{6}{x}^{{2}}+{5}{x}−{1}$$
$$\displaystyle{4}{x}^{{2}}+{4}{x}−{4}$$

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Laaibah Pitt
Given:
$$\displaystyle{6}{x}^{{2}}+{5}{x}-{1}$$
The roots of $$\displaystyle{a}{x}^{{2}}+{b}{x}+{c}$$ are
$$\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}$$
The roots of
$$\displaystyle{6}{x}^{{2}}+{5}{x}-{1}$$
$$\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{5}^{{2}}-{4}{\left({6}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({6}\right)}}}}$$
$$\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{25}+{24}}}}}{{{12}}}}$$
$$\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{49}}}}}{{{12}}}}$$
$$\displaystyle{x}={\frac{{-{5}\pm{7}}}{{{12}}}}$$
$$\displaystyle{x}=-{\frac{{{12}}}{{{12}}}},{\frac{{{2}}}{{{12}}}}$$
$$\displaystyle{x}=-{1},{\frac{{{1}}}{{{6}}}}$$
The roots of $$\displaystyle{6}{x}^{{2}}+{5}{x}-{1}$$ are x=-1,$$\displaystyle{\frac{{{1}}}{{{6}}}}$$
The roots of
$$\displaystyle{4}{x}^{{2}}+{4}{x}-{4}$$
$$\displaystyle{x}={\frac{{-{4}\pm\sqrt{{{4}^{{2}}-{4}{\left({4}\right)}{\left(-{4}\right)}}}}}{{{2}{\left({4}\right)}}}}$$
$$\displaystyle{x}={\frac{{-{4}\pm\sqrt{{{16}+{64}}}}}{{{8}}}}$$
$$\displaystyle{x}={\frac{{-{4}+\sqrt{{{80}}}}}{{{8}}}}$$
$$\displaystyle{x}={\frac{{-{4}\pm{4}\sqrt{{{5}}}}}{{{8}}}}$$
$$\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}+{\frac{{\sqrt{{{5}}}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}-{\frac{{\sqrt{{{5}}}}}{{{2}}}}$$
The roots of $$\displaystyle{4}{x}^{{2}}+{4}{x}-{4}$$ are $$\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}+{\frac{{\sqrt{{{5}}}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}-{\frac{{\sqrt{{{5}}}}}{{{2}}}}$$