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)#