# How do you solve  2x^2 + 3x - 14 = 0?

Jun 30, 2017

2, and $- \frac{7}{2}$

#### Explanation:

$2 {x}^{2} + 3 x - 14 = 0$
Use new Transforming Method.
Transformed equation:
$y ' = {x}^{2} + 3 x - 28 = 0$
Proceeding: find 2 real roots of y', then, divide them by a = 2.
Find 2 real roots, knowing sum (-b = -3) and product (ac = - 28). They are: 4 and -7 (because sum = -3 and product = - 28).
Back to original y, the 2 real roots are:
$x 1 = \frac{4}{a} = \frac{4}{2} = 2$, and $x 2 = - \frac{7}{a} = - \frac{7}{2} = - \frac{7}{2}$