Move all terms to one side.
\[.3{x}^{2}-6x+1-2{x}^{2}+17<0\]
Simplify \(.3{x}^{2}-6x+1-2{x}^{2}+17\) to \(-1.7{x}^{2}-6x+18\).
\[-1.7{x}^{2}-6x+18<0\]
Multiply both sides by \(-1\).
\[1.7{x}^{2}+6x-18>0\]
Factor with quadratic formula.
In general, given \(a{x}^{2}+bx+c\), the factored form is:
\[a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})\]
In this case, \(a=1.7\), \(b=6\) and \(c=-18\).
\[1.7(x-\frac{-6+\sqrt{{6}^{2}-4\times 1.7\times -18}}{2\times 1.7})(x-\frac{-6-\sqrt{{6}^{2}-4\times 1.7\times -18}}{2\times 1.7})\]
Simplify.
\[1.7(x-1.936972)(x+5.466384)\]
\[1.7(x-1.936972)(x+5.466384)>0\]
Solve for \(x\).
Ask: When will \((x-1.936972)(x+5.466384)\) equal zero?
When \(x-1.936972=0\) or \(x+5.466384=0\)
Solve each of the 2 equations above.
\[x=1.936972,-5.466384\]
\[x=1.936972,-5.466384\]
From the values of \(x\) above, we have these 3 intervals to test.
\[\begin{aligned}&x<-5.466384\\&-5.466384<x<1.936972\\&x>1.936972\end{aligned}\]
Pick a test point for each interval.
For the interval \(x<-5.466384\):
Let's pick \(x=-6\). Then, \(.3{(-6)}^{2}-6\times -6+1<2{(-6)}^{2}-17\).After simplifying, we get \(47.8<55\), which is
true
.
Keep this interval.
.
For the interval \(-5.466384<x<1.936972\):
Let's pick \(x=0\). Then, \(.3\times {0}^{2}-6\times 0+1<2\times {0}^{2}-17\).After simplifying, we get \(1<-17\), which is
false
.
Drop this interval.
.
For the interval \(x>1.936972\):
Let's pick \(x=2\). Then, \(.3\times {2}^{2}-6\times 2+1<2\times {2}^{2}-17\).After simplifying, we get \(-9.8<-9\), which is
true
.
Keep this interval.
.
Therefore,
\[x<-5.466384,x>1.936972\]
x<-5.4663841700124;x>1.9369724053065
