[tex]Solution,solve\:for\:d,\:5d^2+4d<3d^2-2d+4\quad :\quad \frac{-3-\sqrt{17}}{2}<d<\frac{\sqrt{17}-3}{2}[/tex]
[tex]Steps:[/tex]
[tex]\mathrm{Subtract\:}3d^2-2d+4\mathrm{\:from\:both\:sides}, 5d^2+4d-\left(3d^2-2d+4\right)<3d^2-2d+4-\left(3d^2-2d+4\right)[/tex]
[tex]\mathrm{Refine}, 2d^2+6d-4<0[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}2, d^2+3d-2<0[/tex]
[tex]d^2+3d-2=0\quad :\quad d=\frac{-3+\sqrt{17}}{2},\:d=\frac{-3-\sqrt{17}}{2}[/tex]
[tex]\mathrm{Factor\:into\:the\:form}\:\left(x-a\right)\left(x-b\right), \left(d-\frac{-3+\sqrt{17}}{2}\right)\left(d-\frac{-3-\sqrt{17}}{2}\right)<0[/tex]
[tex]\mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\left(d-\frac{-3+\sqrt{17}}{2}\right)\left(d-\frac{-3-\sqrt{17}}{2}\right)[/tex]
[tex]\mathrm{Compute\:the\:signs\:of\:}d-\frac{-3+\sqrt{17}}{2}, \mathrm{Compute\:the\:signs\:of\:}d-\frac{-3-\sqrt{17}}{2}[/tex]
[tex]\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:<\:0, \frac{-3-\sqrt{17}}{2}<d<\frac{\sqrt{17}-3}{2}[/tex]
The correct answer is [tex]\frac{-3-\sqrt{17}}{2}<d<\frac{\sqrt{17}-3}{2}[/tex]
