-4=1+y2

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Solution for -4=1+y2 equation: 



-4=1+y2

We move all terms to the left:

-4-(1+y2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-y^2-1-4=0

We add all the numbers together, and all the variables

-1y^2-5=0

a = -1; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·(-1)·(-5)
Δ = -20
Delta is less than zero, so there is no solution for the equation

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