Question 888045
1.

You need to find the critical points of the function. 
Break up the number line according to those critical points and test the inequality in those new regions.
{{{2x^2+x-3>0 }}}
{{{(x-1)(2x+3)>0}}}
In this case the critical points are,
{{{x=1}}} and {{{x=-3/2}}}
So break up the number line,
1.({{{-infinity}}},{{{-3/2}}})
2.({{{-3/2}}},{{{1}}})
3.({{{1}}},{{{infinity}}})
Choose a point in each region and test the inequality.
Region 1: {{{x=-2}}}
{{{2(-2)^2+(-2)-3>0}}}
{{{8-2-3>0}}}
{{{3>0}}}
True
.
.
Region 2: {{{x=0}}}

{{{2(0)^2+(0)-3>0}}}
{{{-3>0}}}
False
.
.
Region 3: {{{x=2}}},
{{{2(2)^2+2-3>0}}}
{{{8+2-3>0}}}
{{{7>0}}}
True
.
.
So the solution region is the union of Region 1 and 3.

({{{-infinity}}},{{{-3/2}}})U({{{1}}},{{{infinity}}})
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.
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2. Convert to vertex form.
{{{y=4x^2-8x+3}}}
{{{y=4(x^2-2x)+3}}}
{{{y=4(x^2-2x+1)+3-4(1)}}}
{{{y=4(x-1)^2-1}}}
Vertex ({{{1}}},{{{-1}}})
Axis : {{{x=1}}}
X-Intercepts: 
{{{4(x-1)^2-1=0}}}
{{{4(x-1)^2=1}}}
{{{(x-1)^2=1/4}}}
{{{x-1=0 +- 1/2}}}
{{{x=1 +- 1/2}}}
{{{x=1/2}}} and {{{x=3/2}}}
Y-intercept:
{{{y=4(0)^2-8(0)+3}}}
{{{y=3}}}
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.
.
{{{graph(300,300,-2,4,-2,4,4x^2-8x+3)}}}
Choose a point not on the parabola.
(0,0) is convenient.
Test the inequality.
{{{y>4x^2-8x+3 }}}
{{{0>4(0)-8(0)+3}}}
{{{0>3}}}
False, so shade the region that doesn't contain (0,0).

{{{graph(300,300,-2,4,-2,4,y>4x^2-8x+3)}}}