# How do you solve 6x^2 - 2 = x ?

${x}_{1} = \frac{2}{3}$
${x}_{2} = - \frac{1}{2}$

#### Explanation:

From the given equation $6 {x}^{2} - 2 = x$

Transpose the x term to the left of the equation

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

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

We can solve this by factoring method

$\left(3 x - 2\right) \left(2 x + 1\right) = 0$

After factoring, equate each factor to 0
First factor equate to 0
$3 x - 2 = 0$

$3 x = 2$

$\frac{3 x}{3} = \frac{2}{3}$

$\frac{\cancel{3} x}{\cancel{3}} = \frac{2}{3}$

$x = \frac{2}{3}$

Second factor equate to 0
$2 x + 1 = 0$
$2 x = - 1$

$\frac{2 x}{2} = \frac{- 1}{2}$

$\frac{\cancel{2} x}{\cancel{2}} = \frac{- 1}{2}$

$x = - \frac{1}{2}$

God bless....I hope the explanation is useful.

Mar 29, 2016

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

#### Explanation:

color(blue)(6x^2-2=x

Subtract $x$ both sides

rarr6x^2-2-x=cancel(x-x

$\rightarrow 6 {x}^{2} - 2 - x = 0$

Rewrite in standard form

rarrcolor(purple)(6x^2-x-2=0

Now,this is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Remember that $a ,$ $b \mathmr{and} c$ are the coeficients of ${x}^{2} ,$ $x \mathmr{and} - 2$

Where

color(red)(a=6,b=-1,c=-2

$\rightarrow x = \frac{- \left(- 1\right) \pm \sqrt{- {1}^{2} - 4 \left(6\right) \left(- 2\right)}}{2 \left(6\right)}$

$\rightarrow x = \frac{1 \pm \sqrt{1 - 4 \left(6\right) \left(- 2\right)}}{12}$

$\rightarrow x = \frac{1 \pm \sqrt{1 - \left(- 48\right)}}{12}$

$\rightarrow x = \frac{1 \pm \sqrt{1 + 48}}{12}$

$\rightarrow x = \frac{1 \pm \sqrt{49}}{12}$

color(orange)(rarrx=(1+-7)/(12)

Now we have $2$ solutions

color(violet)(x=(1+7)/(12)=8/12=2/3

color(indigo)(x=(1-7)/(12)=-6/12=-1/2

color(blue)( :.ul bar |x=2/3,-1/2|