Quadratic Inequalities

First, move everything to the left-hand side: \(4t^2 - 22t + 7 \leq 0\)

 

Now, let's assume that we want to solve \(4t^2 - 22t + 7 = 0\). Using the quadratic formula, this would be \(t = {22 \pm \sqrt{22^2-4 \times 7 \times 4} \over 2 \times 4} ={ 22 \pm 2\sqrt{93} \over 8} = {11 \pm \sqrt{93} \over 4}\)

 

But, we haven't solved the problem yet! We need to find the values that satisfy the inequality. Note that all the points in between the interval of our zeros (\({11 - \sqrt {93} \over 4}\) and \({11 +\sqrt {93} \over 4}\)) will either be positive of negative. 

 

So, let's plug in \(t = 1\) because it's in the interval. This gives us \(4(1)^2 -22 (1) + 7 = -11\), which is less than 0. This means that the value of quadratic must be less than 0 within this interval. 

 

In interval notation, this would be \(\color{brown}\boxed{{11 - \sqrt {93} \over 4}, {11 + \sqrt{93} \over 4}}\)

 

Note: There's probably a better, easier way of solving this that I can't think of.