Areas made from a line and a parabola

First let us calculate the intersection points between the line and the parabola, which I'll call $x_1$ and $x_2$ $(x_1 < x_2)$ :

$\large \displaystyle x^2 + a = \frac32 x$

$\large \displaystyle 2x^2 - 3x + 2a = 0$

$\color{#20A900} \boxed { \large \displaystyle x_1 = \frac{3 - \sqrt{9-16a}}{4} }$

$\color{#20A900} \boxed { \large \displaystyle x_2 = \frac{3 + \sqrt{9-16a}}{4} }$

The areas will be:

$\large \displaystyle S_1 = \int_0^{x_1} \left [ x^2 + a - \frac32 x \right ] dx$

$\color{#20A900} \boxed { \large \displaystyle S_1 = \frac{ (72a-27) + (9-16a)\sqrt{9-16a} }{96} }$

$\large \displaystyle S_2 = \int_{x_1}^{x_2} \left [ \frac32 x - (x^2 + a) \right ] dx$

$\color{#20A900} \boxed { \large \displaystyle S_2 = \frac{(9-16a)\sqrt{9-16a} }{48} }$

Making $S_1 = S_2$ :

$\large \displaystyle 72a-27 + (9-16a)\sqrt{9-16a} = (18-32a)\sqrt{9-16a}$

$\large \displaystyle 72a-27 = (9-16a)\sqrt{9-16a}$

Squaring both sides:

$\large \displaystyle 4096a^3 - 1728a^2 = 0$

Since $a > 0$ :

$\color{#3D99F6} \boxed { \large \displaystyle a = \frac{1728}{4096} = \frac{27}{64} = 0.421875 }$

So, $a$ lies in:

$\color{#3D99F6} \boxed { \large \displaystyle \left ( \frac25 ; \frac{9}{20} \right ) }$