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Sagot :
Answer:
y=-5,y=1
Step-by-step explanation:
To solve this, we need to solve a quadratic equation, using the bhaskara formula.
Initially, let's place the equation in the standard format. So
[tex]2y^2-10y+44=(y-7)^2[/tex][tex]2y^2-10y+44=y^2-14y+49[/tex][tex]2y^2-y^2-10y+14y+44-49=0[/tex][tex]y^2+4y-5=0[/tex]Now we apply the bhaskara formula:
[tex]y=\frac{-(4)\pm\sqrt{4^2-4\ast1\ast-5}}{2\ast1}[/tex]Then
[tex]y=\frac{-4\pm6}{2}[/tex]So two solutions:
[tex]y^{^{\prime}}=\frac{-4+6}{2}=1,y^{^{\prime}^{\prime}}=\frac{-4-6}{2}=-5[/tex]The solution are y=-5,y=1
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