(2y+4)=4y^2+32

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Solution for (2y+4)=4y^2+32 equation: 



(2y+4)=4y^2+32

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4y^2+2y-28=0

a = -4; b = 2; c = -28;
Δ = b2-4ac
Δ = 22-4·(-4)·(-28)
Δ = -444
Delta is less than zero, so there is no solution for the equation

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