You factor a polynomial by writing it as a multiplication of lower-degree polynomials, and going on as long as this is possible.
The quadratic case is quite easy, as it only presents three possibilities: if the discriminant is (striclty) positive you can write p(x)=(x-x_1)(x-x_2), where p(x) is the original quadratic polynomial. If the discriminant equals zero, you have p(x)=(x-x_0)^2. If the discriminant is (strictly) negative, there is no factorization using real numbers.
So, given a quadratic polynomial p(x)=ax^+bx+c, the discriminant \Delta is defined as
\Delta=b^2-4ac
In your case, a=10, b=-18, and c=-4. Plugging these values into the formula, we obtain
\Delta=(-18)^2-4*10*(-4)=324+160=484
We are in the first case, and the solutions x_{1,2} are given by the formula
x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}
And plugging again the values into the formula gives the solutions x_1=-1/5 and x_2=2. Recalling that we have p(x)=(x-x_1)(x-x_2), we factor 10x^2−18x−4 as (x+1/5)(x-2)
