# How do you find the roots for 49x^2-14x-3?

Apr 4, 2018

$- \frac{1}{7} , \mathmr{and} \frac{3}{7}$

#### Explanation:

Use the new Transforming Method (Socratic, Google Search):
$y = 49 {x}^{2} - 14 x - 3 = 0$
Transformed equation:
$y ' = {x}^{2} - 14 x - 147 = 0$
Method: Find 2 real roots of y', then, divide them by a = 49.
Find 2 real roots of y', that have opposite signs, knowing their sum (-b = 14) and their product (ac = - 147). They are -7, and 21.
The 2 real roots of y are:
$x 1 = - \frac{7}{a} = - \frac{7}{49} = - \frac{1}{7}$, and $x 2 = \frac{21}{49} = \frac{3}{7}$