Question

Can you help me please? If there is a solution what’s the solution as well

Answer 1

Given;

`-7y^2+7y+1=0`

Where;

`a=-7,b=7,c=1`

Thus;

`\begin{gathered} b^2-4ac=7^2-4(-7)(1) \\ \\ b^2-4ac>0 \end{gathered}`

Hence, the quadratic equation has two different real solutions.

Then;

`\begin{gathered} y=(-7\pm√(7^2-4(-7)(1)))/(2(-7)) \\ \\ y=1.13,y=-0.13 \end{gathered}`