How do you solve \[4{{x}^{2}}-x+5=0\] ?

Hint: This question is from the topic quadratic equation. For solving this question, we are going to use the formula of discriminant that is \[D={{b}^{2}}-4ac\] , so that we will check if the roots are imaginary or not. If the roots will be imaginary, then we will have to use the value of that is \[i=\sqrt{-1}\]. After that, we will use the formula \[x=\dfrac{-b\pm D}{2a}\], so that we will get the answer.

Complete step by step answer:
Let us solve this question.
This question has asked us to solve the equation \[4{{x}^{2}}-x+5=0\]. We will solve this quadratic equation by using the Sridharacharya method.
Let us understand this method from a general equation.
Let us take a general equation as \[a{{x}^{2}}+bx+c=0\]. Here, the value of a should not be zero and a, b, and c are real.
The formula for discriminant is: \[D={{b}^{2}}-4ac\]
The value of D should be positive or zero for real roots. If the value of D is negative, then the roots will be imaginary (that is there will be negative inside the square root).
And the value of x will be:
\[x=\dfrac{-b\pm \sqrt{D}}{2a}\]
There are two values of x because in between b and D, there is both plus and minus sign.
Now, we will compare the equation \[a{{x}^{2}}+bx+c=0\] with the equation \[4{{x}^{2}}-x+5=0\] and solve it.
Here, we can see after comparing that a=4, b=-1, and c=5.
So, we can write the formula of discriminant and find the value of that to check if the roots are imaginary or not.
\[D={{\left( -1 \right)}^{2}}-4\times 4\times 5=1-80=-79\]
As we can see that the value of D is -79 that is negative. So, we can say that the roots will be imaginary.
Now, by taking square root both sides in the above equation, we get
\[\sqrt{D}=\sqrt{-79}\]
Now, we can see here that negative is inside the root. So, we will use iota here. The symbol of iota is \[i\] and the symbol of iota is \[\sqrt{-1}\]. Or, we can say that \[i=\sqrt{-1}\].
So, we can write the above equation as
\[\sqrt{D}=\sqrt{-1}\times \sqrt{79}=i\sqrt{79}\]
And, the roots of the equation is
\[x=\dfrac{-\left( -1 \right)\pm i\sqrt{79}}{2\times 4}=\dfrac{1\pm i\sqrt{79}}{8}\]
So, we have solved the equation and we have found the values of x.


The values of x are \[\dfrac{1+i\sqrt{79}}{8}\] and \[\dfrac{1-i\sqrt{79}}{8}\].

Note: We should have a better knowledge in the topic quadratic equation to solve this type of question easily. We should know about the Sridharacharya rule for solving this type of question. The Sridharacharya rule says that if the equation is \[a{{x}^{2}}+bx+c=0\], then the roots of the equation will be \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. We should know about iota. The value of iota is the square root of -1 that is \[i=\sqrt{-1}\] .