### Instruction: The graph of the parabola $x = 2y^2 - 6y + 3$ has an $x$-intercept $(a,0)$ and two $y$-intercepts $(0,b)$ and $(0,c)$.  Find $a + b + c$.### Response: To find the $x$-intercept, I need to plug in $y = 0$ and solve for $x$.
So I get $x = 2(0)^2 - 6(0) + 3$, which simplifies to $x = 3$.
Therefore, $a = 3$ and the $x$-intercept is $(3,0)$.
To find the $y$-intercepts, I need to plug in $x = 0$ and solve for $y$.
So I get $0 = 2y^2 - 6y + 3$, which is a quadratic equation.
I can try to factor it, but it doesn't seem to have any nice integer factors.
So I'll use the quadratic formula instead.
The quadratic formula says that $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -6$, and $c = 3$.
Plugging in these values, I get $y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(3)}}{2(2)}$.
Simplifying, I get $y = \frac{6 \pm \sqrt{36 - 24}}{4}$.
Further simplifying, I get $y = \frac{6 \pm \sqrt{12}}{4}$.
I can factor out a 4 from the square root, and get $y = \frac{6 \pm 2\sqrt{3}}{4}$.
I can divide everything by 2, and get $y = \frac{3 \pm \sqrt{3}}{2}$.
These are the two values of $y$ that make $x = 0$, so they are the $y$-intercepts.
Therefore, $b = \frac{3 - \sqrt{3}}{2}$ and $c = \frac{3 + \sqrt{3}}{2}$.
Now I have $a = 3$, $b = \frac{3 - \sqrt{3}}{2}$, and $c = \frac{3 + \sqrt{3}}{2}$.
To find $a + b + c$, I just need to add them up.
I get $a + b + c = 3 + \frac{3 - \sqrt{3}}{2} + \frac{3 + \sqrt{3}}{2}$.
I can combine the fractions with the same denominator, and get $a + b + c = 3 + \frac{6}{2}$.
I can simplify this, and get $a + b + c = 3 + 3 = 6$.