Question

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

Answer 1

Solution:

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;

$\begin{gathered} y=\frac{-7\pm\sqrt{7^2-4(-7)(1)}}{2(-7)} \\ \\ y=1.13,y=-0.13 \end{gathered}$