Question 273993
f(x)=4x^2+7x+2
4x^2+7x+2=0
x=(-7-sqrt(17))/8, x=(-7+sqrt(17)) see below
f(x)<0 ((-7-sqrt(17))/8)<x<((-7+sqrt(17))/8)
f(x)>0 ((-7-sqrt(17))/8)>x>((-7+sqrt(17))/8)
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Ed
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*[invoke quadratic "x", 4, 7, 2]