4-y^2=14y+53

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Solution for 4-y^2=14y+53 equation: 



4-y^2=14y+53

We move all terms to the left:

4-y^2-(14y+53)=0

We add all the numbers together, and all the variables

-1y^2-(14y+53)+4=0

We get rid of parentheses

-1y^2-14y-53+4=0

We add all the numbers together, and all the variables

-1y^2-14y-49=0

a = -1; b = -14; c = -49;
Δ = b2-4ac
Δ = -142-4·(-1)·(-49)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{14}{-2}=-7$

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