Y=(3x2+4x+5)

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Solution for Y=(3x2+4x+5) equation: 



=(3Y^2+4Y+5)

We move all terms to the left:

-((3Y^2+4Y+5))=0



We calculate terms in parentheses: -((3Y^2+4Y+5)), so:

(3Y^2+4Y+5)

We get rid of parentheses

3Y^2+4Y+5

Back to the equation:

-(3Y^2+4Y+5)



We get rid of parentheses

-3Y^2-4Y-5=0

a = -3; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·(-3)·(-5)
Δ = -44
Delta is less than zero, so there is no solution for the equation

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