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

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



(F)=2F^2+2F+5
We move all terms to the left:
(F)-(2F^2+2F+5)=0
We get rid of parentheses
-2F^2+F-2F-5=0
We add all the numbers together, and all the variables
-2F^2-1F-5=0
a = -2; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·(-2)·(-5)
Δ = -39
Delta is less than zero, so there is no solution for the equation

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