# How do you solve 5x^2+19x=4?

Aug 31, 2016

$x = \frac{1}{5} \mathmr{and} x = - 4$

#### Explanation:

As it is a quadratic equation $\Rightarrow$ make = 0

$5 {x}^{2} + 19 x - 4 = 0$

Find factors of 5 and 4 which subtract to make 19.
Signs will be different - more positive.
Cross multiply and subtract

$\textcolor{w h i t e}{\times x} \left(5\right) \text{ } \left(4\right)$

$\textcolor{w h i t e}{\times \times} \textcolor{m a \ge n t a}{5} \text{ } \textcolor{\lim e}{1 \rightarrow 1 \times 1 = 1}$
$\textcolor{w h i t e}{\times \times} \textcolor{\lim e}{1} \text{ "color(magenta)(4 rarr 5xx4=20)" } 20 - 1 = 19$

We have the right factors - now fill in the signs.

$\textcolor{w h i t e}{\times \times} \textcolor{red}{5 \text{ } - 1} \rightarrow 1 \times - 1 = - 1$
color(white)(xxxx)color(blue)(1" "+4) rarr 5xx+4=+20" "+20-1 = +19

$\textcolor{red}{\left(5 x - 1\right)} \textcolor{b l u e}{\left(x + 4\right)} = 0$

Either factor could be equal to 0.

If color(red)(5x-1 = 0, rarr 5x =1 rarr x = 1/5

If color(blue)( x+4=0 rarr x = -4