F(x)=2x^2/5+1

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Solution for F(x)=2x^2/5+1 equation: 



(F)=2F^2/5+1

We move all terms to the left:

(F)-(2F^2/5+1)=0

We get rid of parentheses

-2F^2/5+F-1=0

We multiply all the terms by the denominator


-2F^2+F*5-1*5=0

We add all the numbers together, and all the variables

-2F^2+F*5-5=0

Wy multiply elements

-2F^2+5F-5=0

a = -2; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·(-2)·(-5)
Δ = -15
Delta is less than zero, so there is no solution for the equation

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