Given:
1)
[tex]y^2+y+1=0[/tex]
First find the discriminant,
[tex]\begin{gathered} \text{For ax}^2+bx+c=0 \\ D=b^2-4ac \\ D=1^2-4\times1\times1 \\ D=-3<0 \end{gathered}[/tex]
So, quadratic formula is the best methos to solve this equation.
[tex]\begin{gathered} y^2+y+1=0 \\ y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y=\frac{-1\pm\sqrt[]{1^2-4\times\: 1\times\: 1}}{2\times1} \\ y=\frac{-1\pm\sqrt{3}i}{2} \\ y=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\: y=-\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{gathered}[/tex]
2)
[tex](a-5)^2=9[/tex]
To solve this equation. Take the square root.
So, taking root is the simplest method to solve this equation.
[tex]\begin{gathered} (a-5)^2=9 \\ \text{Take the square root.} \\ a-5=\pm\sqrt[]{9} \\ a-5=\pm3 \\ a-5=3,a-5=-3 \\ a=8,a=2 \end{gathered}[/tex]
