The least positive value of 'a' for which the equation, $$2x^2 + (a - 10)x + \frac{33}{2} = 2a$$ has real roots is

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 $$a=8.$$