H=(2x^2-x+5)+(3x^2-3x-2)

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Solution for H=(2x^2-x+5)+(3x^2-3x-2) equation: 



=(2H^2-H+5)+(3H^2-3H-2)

We move all terms to the left:

-((2H^2-H+5)+(3H^2-3H-2))=0



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

(2H^2-H+5)+(3H^2-3H-2)

We get rid of parentheses

2H^2+3H^2-H-3H+5-2

We add all the numbers together, and all the variables

5H^2-4H+3

Back to the equation:

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



We get rid of parentheses

-5H^2+4H-3=0

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

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