# Factor the quadratic expression 20x^2 + 13x + 2 completely?

A quadratic expression is completely factorizable if and only if its discriminant is positive. Given a quadratic expression of the form $a {x}^{2} + b x + c$, the discriminant $\setminus \Delta$ is defined as ${b}^{2} - 4 a c$. In your case, we have $a = 20$, $b = 13$ and $c = 2$. For this values, we have $\setminus \Delta = 9$, which means that we can factor the expression finding two solutions ${x}_{1}$ and ${x}_{2}$, and thus writing $20 {x}^{2} + 13 x + 2 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a}$.
Since $\setminus \Delta = 9$, its square root equals 3. Plugging the values, we have the two solutions
${x}_{1} = \frac{- 13 + 3}{40} = - \frac{1}{4}$ and
${x}_{2} = \frac{- 13 - 3}{40} = - \frac{2}{5}$.
The factorization is thus $\left(x + \frac{1}{4}\right) \left(x + \frac{2}{5}\right)$