Answer:
[tex]x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i[/tex] are two roots of equation [tex]-3 x^{2}-x-3=0[/tex]
Solution:
Need to solve given equation using quadratic formula.
[tex]-3 x^{2}-x-3=0[/tex]
General form of quadratic equation is [tex]a x^{2}+b x+c=0[/tex]
And quadratic formula for getting roots of quadratic equation is
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
In our case b = -1 , a = -3 and c = -3
Calculating roots of the equation we get
[tex]\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}[/tex]
Since [tex]b^{2}-4 a c[/tex] is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.
[tex]\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}[/tex]
