Answered

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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]

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