<s>[INST] 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?
 [/INST] 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.
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