NOTE: \(\left(x-\dfrac{1}{2}\right)^2=x^2-x+\dfrac{1}{4}\)
\(x^2-x-7=0 \\ \\ \underbrace{x^2-x+\frac{1}{4}}_{\left(x-\frac{1}{2}\right)^2}-7\dfrac{1}{4}=0 \\ \\ \\ \left(x-\frac{1}{2}\right)^2=7\frac{1}{4}=\frac{29}{4} \qquad /\sqrt{}\\ \\ \left|x-\frac{1}{2}\right|=\dfrac{\sqrt{29}}{2} \\ \\ x_1=\dfrac{1}{2}-\dfrac{\sqrt{29}}{2} \approx -2.19 \\ \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{29}}{2} \approx 3.19\)
ANSWER: c)
