For example ,
$$5x^2 - 7x-2=0$$
Then, we only need to find is when $$5x^2-7x=2$$ $$x(5x-7)=2$$ Since that when we get $2-2=0$.
Is there any way we can find them easily without solving quadratics using formula?
I am just getting very enthusiastic in this topic.
Unfortunately, your strategy does not help that much because it is very difficult to solve $$ x(5x-7)=2 $$ without simply returning to the original equation. This is because there is a $2$ on the RHS. The reason we want to be able to write a quadratic in the form $$ (x-p)(x-q)=0 $$ is so that we can exploit the zero-product property: if $a \times b=0$, then $a=0$ or $b=0$. Once we have written an equation in the form $(x-p)(x-q)=0$, we can still instantly deduce that $x=p$ or $x=q$. This is something we can't do when we have something like $$ x(5x-7)=2 \, . $$
Instead, I would suggest dividing through by $5$, which yields $$ x^2 - \frac{7}{5}x - \frac{2}{5} = 0 \, . $$ Then, add $2/5$ to both sides: $$ x^2 - \frac{7}{5}x = \frac{2}{5} \, \tag{*}\label{*} . $$ The LHS is almost a perfect square. Note that $$ \left(x-\frac{7}{10}\right)^2=x^2-\frac{7}{5}x+\frac{49}{100} \, . $$ If we subtract $49/100$ from this equation, we get $$ \left(x-\frac{7}{10}\right)^2-\frac{49}{100}=x^2-\frac{7}{5}x \, . $$ Hence, the equation $\eqref{*}$ can be rewritten as $$ \left(x-\frac{7}{10}\right)^2-\frac{49}{100}=\frac{2}{5} \, . $$ Add $49/100$ to both sides: $$ \left(x-\frac{7}{10}\right)^2 = \frac{2}{5}+\frac{49}{100} = \frac{89}{100} \, . $$ Take the square root of both sides: $$ x-\frac{7}{10} = \pm\sqrt{\frac{89}{100}} \, . $$ Add $7/10$ to both sides: $$ x = \frac{7}{10} \pm \sqrt{\frac{89}{100}} \, , $$ and we are done. This method of solving quadratics is known as completing the square. Unlike factorisation, it can be used to solve any quadratic. In fact, the quadratic formula comes from completing the square on the general quadratic equation $$ ax^2 + bx + c = 0 \, . $$ Please let me know if you have any questions.
All ways of solving quadratic equations are essentially the same, but here's another tack:
You can "depress" the quadratic $ax^2+bx+c$ using the substitution $x=y - b/2a$. This substitution makes the linear term disappear. Then it's easy to solve for $y$ and then plug that answer into the substitution and recover $x$.
For your example: $ 5x^2-7x-2 =0$, let $x = y - \frac{-7}{2\cdot 5} = y+\frac{7}{10}.$ The equation becomes
$$5\left(y+\frac{7}{10}\right)^2 -7\left(y+\frac{7}{10}\right) -2=0$$
which simplifies to
$$5y^2 -\frac{89}{20}=0.$$
Solve for $y$ (which is easy because there's no linear term):
$$y^2=\frac{89}{100}$$
$$y = \pm \frac{\sqrt{89}}{10}.$$
Which means
$$x = \pm \frac{\sqrt{89}}{10} +\frac{7}{10}.$$
So all the steps are easy and obvious as long as you remember the substitution $x = y-b/2a.$
Factoring the equation in the form of $(x-\alpha)(x-\beta)$ is the best way for quadratic equations if the coefficients are reasonably small (not always as in your case because the signs are different). But sometimes (if $\alpha, \beta$ turn out to have an imaginary component or the coefficients are large integers or fractions) the quadratic formula would have to be used. $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ In your case, the quadratic formula would have to be used (or completing the square as @Joe said).
If you want to calculate 'intuitively', there is another method. Note that if $\alpha$ and $\beta$ are the roots of a quadratic of the form $ax^2+bx+c$ ; $a \in \mathbb R - \{0\}$, then $$\alpha + \beta = \frac{-b}{a}$$ and $$\alpha \beta = \frac{c}{a}$$ This can work for equations like $x^2-50x+525$ where $\alpha + \beta = -(-50) = 15+35$ and $\alpha \beta = 525 = 15*35$.
Thus it comes to mind that the roots must be $15$ and $35$ (comparing $LHS$ and $RHS$ due to which the $(x-\alpha)(x-\beta)$ would be $(x-15)(x-35)$.
You need to practice such questions extensively to know which method is more efficient in which equation.
