How do I solve 4x^3-74x^2+300x-250<0?

Your quadratic formula is no help when solving a cubic that has irrational roots.

A graphing calculator can help you get close to the x-intercepts. They are approximately

.. {1.1281, 4.20901, 13.1629}

The inequality is satisfied for

.. x < 1.1281, or

.. 4.20901 < x < 13.1629 ... approximately

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For the nth decimal place, you need to use a machine solver.

There is a factor between 1 and 2 because y(1)=-20 and y(2)=86

x-(4x^3-74x^2+300x-250)/(12x^2-148x+300)

(8x^3-74x^2+250)/(12x^2-148x+300)

Let x1=1... 1.12195, 1.128089, 1.128104, 1.128104

So we can say one factor is (x-1.128104)

Now if you enjoy tediousness you can divide the original equation by the above factor to find the remaining quadratic which is much more easily factored...gotta' love Newton's Approximation :P

4x^3 - 74x^2 + 300x -250

x^2(2x - 37) + 50 (6x +5)

So you've got

(x^2 + 50) (2x-37) (6x+5) < 0

Now solve:

x^2+50 < 0

x^2<-50

x < 5i \sqrt 2

2x-37 <0

2x<37

x<18.5

6x+5<0

6x<-5

x<-5/6