Complete the square to rewrite it as
[tex]2x^2+7x+7=2\left(x^2+\dfrac72x\right)+7=2\left(x^2+\dfrac72x+\left(\dfrac74\right)^2-\left(\dfrac74\right)^2\right)+7[/tex]
[tex]\implies2x^2+7x+7=2\left(x+\dfrac74\right)+\dfrac78[/tex]
The minimum value of this expression occurs at [tex]x=-\dfrac74[/tex] because [tex]\left(x+\dfrac74\right)^2[/tex] will always be non-negative. Then at this point we get a value of [tex]\dfrac78[/tex], which means
[tex]2x^2+7x+7\ge\dfrac78[/tex]
and so is always positive for any real value of [tex]x[/tex]
