Quadratic inequation

If 5x + x² > 100, then x is not

Correct result:

x =  0

Solution:

5x+x2100=0  5x+x2100=0 x2+5x100=0  a=1;b=5;c=100 D=b24ac=5241(100)=425 D>0  x1,2=b±D2a=5±4252=5±5172 x1,2=2.5±10.307764064 x1=7.80776406404 x2=12.807764064   Factored form of the equation:  (x7.80776406404)(x+12.807764064)=0 x>x1 x=x<x2=05x+x^2-100=0 \ \\ \ \\ 5x+x^2-100=0 \ \\ x^2 +5x -100=0 \ \\ \ \\ a=1; b=5; c=-100 \ \\ D=b^2 - 4ac=5^2 - 4\cdot 1 \cdot (-100)=425 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -5 \pm \sqrt{ 425 } }{ 2 }=\dfrac{ -5 \pm 5 \sqrt{ 17 } }{ 2 } \ \\ x_{1,2}=-2.5 \pm 10.307764064 \ \\ x_{1}=7.80776406404 \ \\ x_{2}=-12.807764064 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -7.80776406404) (x +12.807764064)=0 \ \\ x>x_{1} \ \\ x=x<x_{2}=0

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x not in interval <-12.807764064; 7.80776406404>

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