If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$? 
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?

The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
 A: You used the quadratic formula incorrectly.
$$\begin{align*}20x^2 + 13x - 15 = 0 
\implies
x &= \frac{-(13) \pm \sqrt{(13)^2 - 4(20)(-15)}}{2(20)}\\
&= \frac{-13 \pm \sqrt{169 + 1200}}{40} \\
&= \frac{-13 \pm \sqrt{1369}}{40} \\
&= \frac{-13 \pm 37}{40} \\
&= \frac{24}{40} \text{ or } - \frac{50}{40} \\
&= \frac 3 5 \text{ or } - \frac 5 4 \\
\end{align*}$$
Therefore
$$ 20 x^2 + 13 x - 15 = 20 \left( x - \frac 3 5 \right) \left( x + \frac 5 4 \right)$$
Bring a factor of $5$ into the first parenthetical and $4$ into the second, each from the $20$, to get the desired form and find your answer.
A: If the question is asking you to factor you need not necessarily solve for the roots.
Use the FOIL (First, Out, In, Last) method to expand $(ax + b)(cx + d)$ and simultaneously solve for the coefficients $a, b, c, d$.
We would have $$\begin{cases} ac = 20 \\ ad + bc = 13 \\ bd = -15 \end{cases}$$
Furthermore, once you become more familiar and proficient with FOIL, you can try to guess factors and evaluate cases to see which $2$ factors of $20$ and $2$ factors of $-15$ add up to $13$ directly. This would be as such:
$20 \to \{\pm1, \pm2, \pm4, \pm5, \pm10, \pm20\}$
$-15 \to \{\pm1, \pm3, \mp 5, \mp15\}$
See that $5 \cdot 5 + 4 \cdot -3 = 13$; this is exactly in the form $ad + bc$. Then, $5 \cdot 4 = 20$ and $5 \cdot -3 = -15$, as required.
A: Another way to realize the solution (through inspection):
\begin{align*}
20x^{2} + 13x - 15 & = (20x^{2} - 12x) + (25x - 15)\\\\
& = 4x(5x - 3) + 5(5x - 3)\\\\
& = (4x + 5)(5x - 3)
\end{align*}
