How to solve this problem?

4a² - 10a + 6  =  0       First, let's divide through by  2  .

2a² - 5a + 3  =  0

We can use the Quadratic formula to solve this problem for  a  .

The Quadratic formula says, if you have an equation written in the form

Ax² + Bx + C  =  0   , then...   $x = {-{\color{blue}B} \pm \sqrt{{\color{blue}B}^2-4{\color{red}A}{\color{green}C}} \over 2{\color{red}A}}$

So, for our problem...

2a² + -5a + 3  =  0

x = a ,  A = 2 ,  B = -5 ,  C = 3  .    Substitute these values into the Quadratic formula.

$a = {-{\color{blue}(-5)} \pm \sqrt{{\color{blue}(-5)}^2-4{\color{red}(2)}{\color{green}(3)}} \over 2{\color{red}(2)}} \\~\\ a = {5 \pm \sqrt{25-24} \over 4} \\~\\ a={5 \pm \sqrt{1} \over 4} \\~\\ a={5 \pm 1 \over 4} \\~\\ \begin{array}\ a=\frac{5+1}{4}\qquad \text{ or }\qquad &a&=\frac{5-1}{4} \\~\\ a=\frac{6}{4}&a&=\frac44 \\~\\ a=\frac32&a&=1 \\~\\ \end{array}$

Hectictar's method is perfectly executed, but this equation is also factorable. How do I know? There is a universal test that determines if a quadratic is factorable. It is the following:

If $\sqrt{b^2-4ac}\in\mathbb{Q}\hspace{1mm}\text{and}\sqrt{b^2-4ac}\geq0$, then the quadratic is factorable. In other words, the standard form of a quadratic equation is $ax^2+bx+c$. If you apply the rule above and get a rational number (integers and decimals that terminate or repeat) and is greater than or equal to 0, then the quadratic is factorable. Let's try it:

$4a^2-10a+6=0$

Before, we start, let's determine our a's, b's, and c's. Our a is the coefficient of the quadratic term. In this case, that is 4. OUr b is the coefficient in front of the linear term, which is -10. Make sure to include the sign when doing this calculation. Our c is our constant term, which is 6. Let's apply the rule and see if this equation is factorable:
 

Great! Now that we have determined that this quadratic can be factored, let's factor it! I'll use a picture to demonstrate what I am doing:
 

Our job is to find 2 factors that both multiply to get -24 and add to get -10. If you toy with the numbers for some time, you will eventually figure this out:

Now that we have figured out which two numbers satisfy the above problem, let's split the b-term:

Therefore, your solution set is the following:

$a_1=\frac{3}{2}$

$a_2=1$

Now, you are done!

We can just factorize it :)

4a^2 - 10a + 6 = 0

2a^2 - 5a + 3 = 0

(2a - something)(a - something)=0

We don't know what it is yet, but think of 2 numbers that satisfies:

$-2x-y=-5\\ xy=3$

We can easily see that x = 1 and y = 3.

(2a - 3)(a - 1) = 0

a = 3/2 or a = 1 :)