# How do you solve 3y²+14y=5?

Mar 27, 2016

$\frac{1}{3} \mathmr{and} - 5$

#### Explanation:

$f \left(y\right) = 3 {y}^{2} + 14 y = 5 = 0.$
Use the new Transforming Method (Google, Yahoo Search).
Transformed equation: $f ' \left(y\right) = {y}^{2} + 14 y - 15.$
Since a + b + c = 0, the 2 real roots of f'(y) are: ( 1) and $\left(\frac{c}{a} = - 15\right)$.
Therefor, the 2 real roots of f(y) are: $\left(\frac{1}{3}\right) \mathmr{and} \left(- \frac{15}{3} = - 5\right)$
Answers: $\frac{1}{3}$ and -5