H=(3x^2+13)

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Solution for H=(3x^2+13) equation: 



=(3H^2+13)

We move all terms to the left:

-((3H^2+13))=0



We calculate terms in parentheses: -((3H^2+13)), so:

(3H^2+13)

We get rid of parentheses

3H^2+13

Back to the equation:

-(3H^2+13)



We get rid of parentheses

-3H^2-13=0

a = -3; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·(-3)·(-13)
Δ = -156
Delta is less than zero, so there is no solution for the equation

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