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Answer:
The zeroes of this polynomial are [tex]x_{1} \approx 0.847[/tex] and [tex]x_{2} \approx -1.180[/tex].
Step-by-step explanation:
Let [tex]3\cdot x^{2}+x - 3 = 0[/tex], the quickest and most efficient approach to find the zeroes of this second order polynomial is by Quadratic Formula. For all [tex]a\cdot x^{2}+b\cdot x + c = 0[/tex], roots are determined by:
[tex]x_{1,2} = \frac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}[/tex] (1)
Where [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are coefficients of the polynomial.
If we know that [tex]a = 3[/tex], [tex]b = 1[/tex] and [tex]c = -3[/tex], then roots of the polynomial are, respectively:
[tex]x_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (3)\cdot (-3)}}{2\cdot (3)}[/tex]
[tex]x_{1,2} =\frac{-1\pm \sqrt{37}}{6}[/tex]
The zeroes of this polynomial are [tex]x_{1} \approx 0.847[/tex] and [tex]x_{2} \approx -1.180[/tex].