Consider the given quadratic equation,
[tex]x^2+x=\frac{19}{4}[/tex]Here the coefficient of 'x' is positive, so the algebraic identity which will be used in the method of completing squares, is as follows,
[tex]a^2+2ab+b^2=\mleft(a+b\mright)^2[/tex]Writing the left side of the given equation in the form of the left side expression of the identity,
[tex](x)^2+2(x)(\frac{1}{2})=\frac{19}{4}[/tex]It is observed that,
[tex]\begin{gathered} a=x \\ b=\frac{1}{2} \end{gathered}[/tex]Note that, in order to complete the square, the left side expression requires a 'b-squared' term.
This can be done by adding the term to both sides of the equality.
Adding the term both sides,
[tex]\begin{gathered} (x)^2+2(x)(\frac{1}{2})+(\frac{1}{2})^2=\frac{19}{4}+(\frac{1}{2})^2 \\ (x)^2+2(x)(\frac{1}{2})+(\frac{1}{2})^2=\frac{19}{4}+(\frac{1}{4})^{} \\ (x)^2+2(x)(\frac{1}{2})+(\frac{1}{2})^2=\frac{19+1}{4} \\ (x)^2+2(x)(\frac{1}{2})+(\frac{1}{2})^2=5 \end{gathered}[/tex]Now, the left side of the expression perfectly fits the left side expression of the algebraic identity, which can be replaced by its right side term,
[tex](x+\frac{1}{2})^2=5[/tex]Taking square rootsboth sides,
[tex]\begin{gathered} x+\frac{1}{2}=\pm\sqrt[]{5} \\ x+\frac{1}{2}=\sqrt[]{5}\text{ }or\text{ }x+\frac{1}{2}=-\sqrt[]{5}\text{ } \\ x=\sqrt[]{5}-\frac{1}{2}or\text{ }x=-\sqrt[]{5}-\frac{1}{2} \end{gathered}[/tex]Thus, the solutions of the given quadratic equations are,
[tex]x=\sqrt[]{5}-\frac{1}{2}or\text{ }x=-\sqrt[]{5}-\frac{1}{2}[/tex]Therefore, option C is the correct choice.
