(y+2)(y+2)=2y^2+7y+32

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Solution for (y+2)(y+2)=2y^2+7y+32 equation: 



(y+2)(y+2)=2y^2+7y+32

We move all terms to the left:

(y+2)(y+2)-(2y^2+7y+32)=0

We get rid of parentheses

-2y^2+(y+2)(y+2)-7y-32=0

We multiply parentheses ..

-2y^2+(+y^2+2y+2y+4)-7y-32=0

We get rid of parentheses

-2y^2+y^2+2y+2y-7y+4-32=0

We add all the numbers together, and all the variables

-1y^2-3y-28=0

a = -1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·(-1)·(-28)
Δ = -103
Delta is less than zero, so there is no solution for the equation

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