Find the minimum value!

Thank you Mrs.Julian Poon! I love your solution! And here is my solution:

We can convert our equation into this:

$6{ x }^{ 2 }+7x+5$

= $6{ x }^{ 2 }+2\times \sqrt { 6 } \times \frac { 7\sqrt { 6 } }{ 12 } x+({ \frac { 7\sqrt { 6 } }{ 12 } }^{ 2 })+\frac { 71 }{ 24 }$

= ${ (\sqrt { 6 } x-\frac { 7\sqrt { 6 } }{ 12 } ) }^{ 2 }+\frac { 71 }{ 24 }$

Because ${ \left( \sqrt { 6 } x-\frac { 7\sqrt { 6 } }{ 12 } \right) }^{ 2 }\ge 0$ with all values of x.

So the equation reaches the value of N if ${ \left( \sqrt { 6 } x-\frac { 7\sqrt { 6 } }{ 12 } \right) }^{ 2 }$ reaches its minimum value, and its minimum value is 0

So N is $\frac { 71 }{ 24 }$

And $\left\lfloor N \right\rfloor =2$

$6{ x }^{ 2 }+7x+5=y$ would give a quadratic graph with a min point but no max point. To find the min of this equation, find the min of $x$

Substituting: $y=0$ makes $x=$ no real solution.

Substituting: $y=10$ makes $x=$ $\frac { 1 }{ 2 }$ or $-\frac { 5 }{ 3 }$

Therefore, the min value of $x=0.5\left[ -\frac { 5 }{ 3 } +\frac { 1 }{ 2 } \right] =-\frac { 7 }{ 12 }$

Substituting back into the equation gives, $\frac { 71 }{ 24 }$

So, $\left\lfloor \frac { 71 }{ 24 } \right\rfloor =\boxed { 2 }$

This is a parabola; it's symmetric. When set to zero it has no real roots. Let me subtract an arbitrary number from the left side of the equation so that it does have real roots. Since the parabola is symmetric, the minimum will be at the midpoint of those roots: -7/12. Now it's just a matter of evaluating the given polynomial at that point.

The minimum/maximum point of the quadratic graphic is given by (x,y) = (-b/2a , (-b²+4ac)/4a). So the minimum y will be: (-49+4 6 5)/(4*6) = 71/24 = 2,95. The integer part is 2.

y=6x^2+7x+5 dy/dx=12x+7=0 x=-7/12 so y=2.9 [N]=2