Solve the given quadratic equation $3{{x}^{2}}+11x+10=0$.

Hint: Consider the given quadratic $3{{x}^{2}}+11x+10$ and write it in its factors form as, $3{{x}^{2}}+6x+5x+10=0$ and then factorised as $\left( 3x+5 \right)\left( x+2 \right)=0$ and finally get value of x.

Complete step-by-step solution:
We are given a quadratic equation $3{{x}^{2}}+11x+10=0$ and we have to solve to find the value of x.
$3{{x}^{2}}+11x+10=0$ Is consider as quadratic equation is any equation that can be rearranged in standard form as $a{{x}^{2}}+bx+c=0$ .
Here x represents unknown and a, b, c are known numbers where $a\ne 0$. Otherwise, it becomes linear as no $a{{x}^{2}}$ term is there. The number a, b, c are coefficients of the equation and may be distinguished by calling them respectively, the quadratic coefficient, linear coefficient, and the constant or free term.
The values of x that satisfy the equation are called a solution of the equation and roots or zeroes of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is no real solution, one says it is double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation of the form $a{{x}^{2}}+bx+c=0$ can be factored as $a\left( x-\alpha \right)\left( x-\beta \right)=0$ where $\alpha $ and $\beta $ are solution of x.
Because the quadratic equation involves only one known, it is called coordinate. The quadratic equation only contains the power of x that are non – negative integers, and therefore it is a polynomial equation. In particular, it is a second-degree polynomial equation.
As the equation given is $3{{x}^{2}}+11x+10=0$
 So, $3{{x}^{2}}+11x+10=0$ .
Can be written as,
$\Rightarrow 3{{x}^{2}}+6x+5x+10=0$.
$\Rightarrow 3(x+2)+5(x+2)=0$.
Which can be factorised as,
 $\left( x+2 \right)\left( 3x+5 \right)=0$ .
So, for the equation value, $-2$ and $\dfrac{-5}{3}$ satisfies equation in place of x.
Hence the values of x are $-2$ and $\dfrac{-5}{3}$.

Note: We can also solve it by another method by using a formula directly which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{\text{2a}}$ .
If the given quadratic equation is $a{{x}^{2}}+bx+c=0$ and the variable is x which needs to be found out. But this will take more calculations and time so there is another method
$3{{x}^{2}}+11x+10=0$
We will divide this equation by 3 to make perfect square
 ${{x}^{2}}+\dfrac{11}{3}x+\dfrac{10}{3}=0$
${{x}^{2}}+2.x.\dfrac{11}{3\times 2}+{{\left( \dfrac{11}{6} \right)}^{2}}={{\left( \dfrac{11}{6} \right)}^{2}}-\dfrac{10}{3}$
${{\left( x+\dfrac{11}{6} \right)}^{2}}=\dfrac{121}{36}-\dfrac{10}{3}$
${{\left( x+\dfrac{11}{6} \right)}^{2}}=\dfrac{121-120}{36}$
${{\left( x+\dfrac{11}{6} \right)}^{2}}=\dfrac{1}{36}$
Take square root on both side
$x+\dfrac{11}{6}=\pm \dfrac{1}{6}$
$x=-\dfrac{11}{6}+\dfrac{1}{6}\,\,or\,x=-\dfrac{11}{6}-\dfrac{1}{6}$
$x=\dfrac{-11+1}{6}\,\,or\,x=\dfrac{-11-1}{6}$
$x=\dfrac{-10}{6}\,\,or\,x=\dfrac{-12}{6}$
$x=\dfrac{-5}{3}\,\,or\,x=-2$