Get comprehensive solutions to your questions with the help of IDNLearn.com's experts. Discover in-depth and trustworthy answers to all your questions from our experienced community members.

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

Can You Help Me Please If There Is A Solution Whats The Solution As Well class=

Sagot :

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]

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.