How do you solve the equation $2{{x}^{2}}+x+7=0$ ?

Hint: Now we are given with a quadratic equation in x. Now first we will compare the given equation with the general quadratic equation $a{{x}^{2}}+bx+c$ and then find the values of a, b and c. Now we know the roots of the quadratic equation is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ hence we will substitute the values of a, b and c in the formula and find the roots of the equation.

Complete step by step solution:
Now consider the given equation $2{{x}^{2}}+x+7=0$
Now since the degree of the equation is 2 we can say the equation is a quadratic equation in x of the form $a{{x}^{2}}+bx+c$ where a = 2, b = 1 and c = 7.
Now we want to find the solution of the equation which means we want to find the roots of the equation.
Now we know that the roots of the quadratic equation of the form $a{{x}^{2}}+bx+c$ are given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Hence on substituting the values of a, b and c in the equation we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 2 \right)\left( 7 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{1-56}}{4} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{-55}}{4} \\
\end{align}$
Now we know that $\sqrt{-1}=i$ hence using this we get,
$\Rightarrow x=\dfrac{-1\pm i\sqrt{55}}{4}$

Hence the solution of the given quadratic equation is $x=\dfrac{-1+i\sqrt{55}}{4}$ and $x=\dfrac{-1-i\sqrt{55}}{4}$ .

Note: Now note that for any quadratic equation of the form $a{{x}^{2}}+bx+c$ if we have ${{b}^{2}}-4ac>0$ then the roots are real and distinct. If we have ${{b}^{2}}-4ac=0$ then the roots are real and repeating. But if the value of ${{b}^{2}}-4ac<0$ then the roots are complex roots which are conjugate with each other. This value ${{b}^{2}}-4ac$ is called the discriminant of the quadratic equation and is denoted by D. Hence discriminant determines the nature of roots.