# How do you solve 6x² - 4 = 5x?

Jul 29, 2015

$x = - \frac{1}{2} , \frac{4}{3}$

#### Explanation:

$6 {x}^{2} - 4 = 5 x$

Gather all terms on the left side.

$6 {x}^{2} - 5 x - 4 = 0$

This is a quadratic equation, $a {x}^{2} + b x + c$, where $a = 6 , b = - 5 , \mathmr{and} c = - 4$.

You can use the quadratic formula to solve this equation.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 5\right) \pm \sqrt{- {5}^{2} - 4 \cdot 6 \cdot - 4}}{2 \cdot 6}$ =

$x = \frac{5 \pm \sqrt{25 + 96}}{12}$ =

$x = \frac{5 \pm \sqrt{121}}{12}$ =

$x = \frac{5 \pm 11}{12}$

Solve for $x$.

$x = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3}$

$x = \frac{5 - 11}{12} = - \frac{6}{12} = - \frac{1}{2}$

$x = - \frac{1}{2} , \frac{4}{3}$

Jul 30, 2015

Solve y = 6x^2 - 5x - 4 = 0 (1)
Ans: - 1/2 and 4/3

#### Explanation:

I use the new Transforming Method.
Transformed y' = x^2 - 5x - 24. (2). Roots have opposite signs.
Factor pairs of (-24) -> (-2, 12)(-3, 8). This sum is 5 = -b. Two real roots of (2) are: y1 = -3 and y2 = 8.
Back to original equation (1), the 2 real roots are:
$x 1 = y \frac{1}{a} = - \frac{3}{6} = - \frac{1}{2}$ and $x 2 = y \frac{2}{a} = \frac{8}{6} = \frac{4}{3}$