How do you solve $6 x^{2} + 2 x = 104$ $6x^2 + 2x = 104$ ?

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How do you solve $6 x^{2} + 2 x = 104$ $6x^2 + 2x = 104$ ?

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Answer 1 · 2018-04-01T22:04:32

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

Explanation:

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

Apply the quadratic formula.

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

$x = \frac{- 2 \pm \sqrt 2^{2} - 4 \cdot 6 \cdot (- 104)}{2 \cdot 6}$ $x=(-2±√2^2-4*6*(-104))/(2*6)$

Simplify to get
$x = \frac{- 2 \pm \sqrt 2^{2} + 2496}{12}$ $x=(-2±√2^2+2496)/12$

Simplify even more to get $x = \frac{- 2 \pm \sqrt 2500}{12}$ $x=(-2±√2500)/12$
$x = \frac{- 2 \pm 50}{12}$ $x=(-2±50)/12$

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

Yay!

Answer 2 · 2018-04-01T23:10:45

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

Explanation:

Use the new Transforming Method (Socratic, Google Search).
$6 x^{2} + 2 x - 104 = 0$ $6x^2 + 2x - 104 = 0$
$y = 3 x^{2} + x - 52 = 0$ $y = 3x^2 + x - 52 = 0$
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:
$x1 = 12/a = 12/3 = 4$ , and $x2 = - 13/a = - 13/3$

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How do you solve $6 x^{2} + 2 x = 104$ $6x^2 + 2x = 104$ ?

2 Answers

Explanation:

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