Solution:
Given;
[tex]-7y^2+7y+1=0[/tex]
Where;
[tex]a=-7,b=7,c=1[/tex]
Thus;
[tex]\begin{gathered} b^2-4ac=7^2-4(-7)(1) \\ \\ b^2-4ac>0 \end{gathered}[/tex]
Hence, the quadratic equation has two different real solutions.
Then;
[tex]\begin{gathered} y=\frac{-7\pm\sqrt{7^2-4(-7)(1)}}{2(-7)} \\ \\ y=1.13,y=-0.13 \end{gathered}[/tex]
