Y=2c^2-16c+32

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Solution for Y=2c^2-16c+32 equation: 



=2Y^2-16Y+32

We move all terms to the left:

-(2Y^2-16Y+32)=0

We get rid of parentheses

-2Y^2+16Y-32=0

a = -2; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·(-2)·(-32)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$Y=\frac{-b}{2a}=\frac{-16}{-4}=+4$

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