40y=25y^2+16

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Solution for 40y=25y^2+16 equation: 



40y=25y^2+16

We move all terms to the left:

40y-(25y^2+16)=0

We get rid of parentheses

-25y^2+40y-16=0

a = -25; b = 40; c = -16;
Δ = b2-4ac
Δ = 402-4·(-25)·(-16)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{-40}{-50}=4/5$

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