Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Given equation,

${{x}^{2}}+\frac{11}{3}+\frac{10}{3}=0$

${{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0$

${{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0$

${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}$

${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}$

${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121-120}{36}$

${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}$

${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}$

$x+\frac{11}{6}=\pm \frac{1}{6}$

$x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}$

$x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}$

$x=\frac{-10}{6}orx=\frac{-12}{6}=-2$

${{x}^{2}}+\frac{11x}{3}+\frac{10}{3}=0$

${{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0$

${{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0$

${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}$

${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}$

${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}$

${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}$

$x+\frac{11}{6}=\pm \frac{1}{6}$

$x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}$

$x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}$

$x=\frac{-10}{6}oex=\frac{-12}{6}=-2$

⇒ $x=-5/3=3$ or $x=-2$

Thus, the roots of the given quadratic equation are $x=-5/3$ and $x=-2.$