The question is 2x^2 - 7x + 3 = 0.
First of all quadratic equation is written in the form of ax^2 + bx + c = 0 in where a,b and c are numbers. In this equation I was told to use the matrix method. Since this equation doesnt have a common factor, I have to factorise the equation. The thing is I'm not really sure where to start
$2x^2 - 7x + 3 = 0$
To split the linear term of $2x^2 - 7x + 3$, we must find two numbers with product $2 \cdot 3 = 6$ and sum $-7$. They are $-6$ and $-1$. Hence,
\begin{align*} 2x^2 - 6x - x + 3 & = 0 && \text{split the linear term}\\ 2x(x - 3) - 1(x - 3) & = 0 && \text{factor by grouping}\\ (2x - 1)(x - 3) & = 0 && \text{extract the common factor} \end{align*}
If a product is equal to zero, then one of the factors must equal zero. Thus, $2x - 1 = 0$ or $x - 3 = 0$. Solve each linear equation for $x$, then check your answers by substituting them into the equation $2x^2 - 7x + 3 = 0$.
Note that $$\begin{array}{l}2{x^2} - 7x + 3 = 0\\{x^2} - \frac{7}{2}x + \frac{3}{2} = 0\\{\left( {x - \frac{7}{4}} \right)^2} - \frac{{49}}{4} + \frac{3}{2} = 0\end{array}$$could you see the answer?
By the rational root theorem, if $2x^2-7x+3=0$ has a rational root, then it has to be either of $$\pm 1,\pm\frac 12,\pm 3,\pm\frac 32.$$
Since $2\cdot 3^2-7\times 3+3=0$, by the factor theorem, we have $$2x^2-7x+3=(x-3) (2 x-1).$$
