# How do you solve 6x^2 + 2x = 104?

Apr 1, 2018

$x = - 4 , - \frac{13}{3}$

#### Explanation:

Subtract $104$ from both sides to get
$6 {x}^{2} + 2 x - 104 = 104 - 104$ => $6 {x}^{2} + 2 x - 104 = 0$.

$a = 6$
$b = 2$
$c = - 104$

x=(-2±√2^2-4*6*(-104))/(2*6)

Simplify to get
x=(-2±√2^2+2496)/12

Simplify even more to get x=(-2±√2500)/12
x=(-2±50)/12

If $x = \frac{- 2 + 50}{12}$ then $x = - 4$.
If $x = \frac{- 2 - 50}{12}$ then $x = - \frac{13}{3}$

Yay!

Apr 1, 2018

4 and $- \frac{13}{3}$

#### Explanation:

Use the new Transforming Method (Socratic, Google Search).
$6 {x}^{2} + 2 x - 104 = 0$
$y = 3 {x}^{2} + x - 52 = 0$
Transformed equation:
$y ' = {x}^{2} + x - 156 = 0$
Proceeding. Find the 2 real roots of y', then, divide them by a = 3.
Find 2 real roots, that have opposite signs (ac < 0), knowing their sum (-b = -1) and their product (ac = - 156). They are 12 and - 13.
Back to y, the 2 real roots of y are:
$x 1 = \frac{12}{a} = \frac{12}{3} = 4$, and $x 2 = - \frac{13}{a} = - \frac{13}{3}$