# How do you solve 6x^2 - x -3 =0?

Nov 22, 2015

$x \approx - 0.63 , \textcolor{w h i t e}{\times} x \approx 0.79$

#### Explanation:

It is a matter of practice so that you get used to spotting the factors.

The only 2 factors you are going to get for 3 is 1 and 3. This is because 3 is a prime number. The thing is that we have -3 so one has to be positive and the other negative.
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Factors of 3 are only {1,3} because 3 is a prime number

Factors of 6 are {1,6} , {2,3}

Looking for -1 in $- x$

These factors do not work so revert to the standard solution formula:

$a {x}^{2} + b x + c = 0$

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

$a = 6$
$b = - 1$
$c = - 3$

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(6\right) \left(- 3\right)}}{2 \left(6\right)}$

$\textcolor{g r e e n}{\text{The use of brackets round negative numbers reduce}}$
$\textcolor{g r e e n}{\text{the chance of making a mistake!}}$

$x = \frac{1 \pm \sqrt{1 + 72}}{12}$

$x = \frac{1}{12} \pm \frac{\sqrt{73}}{12}$

$x \approx 0.08 \dot{3} \pm 0.71$

$x \approx - 0.63 , x \approx 0.79$