We have: $$2x^{2}+(a-10)x+\dfrac{33}{2}=2a,$$
Or, $$2x^{2}+(a-10)x+\dfrac{33}{2}-2a=0.$$
We get: $$A=2,\qquad B=a-10,\qquad C=\dfrac{33}{2}-2a.$$
For $$Ax^{2}+Bx+C=0,$$ the discriminant is $$\Delta=B^{2}-4AC,$$ and real roots exist when $$\Delta\ge0.$$
Applying this, we write
$$\Delta=(a-10)^{2}-4\cdot2\left(\dfrac{33}{2}-2a\right)\ge0.$$
$$(a-10)^{2}=a^{2}-20a+100.$$
Next, evaluate the product in the second term:
$$4\cdot2=8,$$
so $$4AC=8\left(\dfrac{33}{2}-2a\right).$$
$$8\left(\dfrac{33}{2}\right)-8(2a)=8\cdot\dfrac{33}{2}-16a=4\cdot33-16a=132-16a.$$
Putting it back in the discriminant,
$$a^{2}-20a+100-\left(132-16a\right)\ge0.$$
$$a^{2}-20a+100-132+16a\ge0.$$
For the $$a$$ terms: $$-20a+16a=-4a.$$
For the constant terms: $$100-132=-32.$$
Hence, we get: $$a^{2}-4a-32\ge0.$$
To see where this inequality holds, we first solve the corresponding quadratic equation
$$a^{2}-4a-32=0.$$
Using the quadratic formula $$a=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\;b=-4,\;c=-32,$$ we write
$$a=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-32)}}{2(1)}=\dfrac{4\pm\sqrt{16+128}}{2}.$$
Simplify under the square root:
$$16+128=144,\qquad\sqrt{144}=12.$$
Therefore,
$$a=\dfrac{4\pm12}{2}.$$
This gives two real numbers:
$$a_{1}=\dfrac{4+12}{2}=\dfrac{16}{2}=8,\qquad
a_{2}=\dfrac{4-12}{2}=\dfrac{-8}{2}=-4.$$
The quadratic $$a^{2}-4a-32$$ has a positive leading coefficient, so its graph is an upward-opening parabola. Hence the inequality $$a^{2}-4a-32\ge0$$ is satisfied outside the interval between its roots, that is, for
$$a\le-4\quad\text{or}\quad a\ge8.$$
Hence, the least positive value of $$a=8.$$
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