# What are the solutions to 5x^2+27x+10=0?

Jun 15, 2017

$x = - 5 \text{ or } x = - \frac{2}{5}$

#### Explanation:

$\text{factorise by 'splitting' the term in x}$

$\Rightarrow 5 {x}^{2} + 25 x + 2 x + 10 = 0 \leftarrow 25 x + 2 x = 27 x$

$\Rightarrow \textcolor{red}{5 x} \left(x + 5\right) + \textcolor{red}{2} \left(x + 5\right) = 0$

$\Rightarrow \left(x + 5\right) \left(\textcolor{red}{5 x + 2}\right) = 0$

$\text{equating each factor to zero}$

$\Rightarrow x + 5 = 0 \Rightarrow x = - 5$

$5 x + 2 = 0 \Rightarrow x = - \frac{2}{5}$

Jun 16, 2017

$- 5$ and $- \frac{2}{5}$

#### Explanation:

Solving by the new Transforming Method (Google, Socratic Search).
$y = 5 {x}^{2} + 27 x + 10 = 0$
Transformed equation:
$y ' = {x}^{2} + 27 x + 50 = 0$--> (ac = 50)
Proceeding: Find 2 real roots of y', then, divide them by a = 5.
Find 2 real roots knowing sum (-b = -27) and product (c = 50). They are -2 and -25.
Back to y, the 2 real roots are:
$- \frac{2}{a} = - \frac{2}{5}$, and $- \frac{25}{a} = - \frac{25}{5} = - 5$
NOTE:
There are no factoring by grouping and no solving the 2 binomials.