Can someone please help me out with this? i am trying to solve: 54= l^2- l3

Jun 3, 2017

The solutions are $S = \left\{- 6 , 9\right\}$

Explanation:

$54 = {l}^{2} - l 3$

We rewrite it as

${l}^{2} - 3 l - 54 = 0$

We solve this by factorising

$\left(l + 6\right) \left(l - 9\right) = 0$

The solutions are

$l + 6 = 0$, $\implies$, $l = - 6$

and

$l - 9 = 0$, $\implies$, $l = 9$

Jun 3, 2017

$l = 9 , \mathmr{and} , l = - 6.$

Explanation:

${l}^{2} - 3 l = 54$

$\Rightarrow {l}^{2} - 3 l - 54 = 0.$

Now, $9 \times 6 = 54 , \mathmr{and} , 9 - 6 = 3.$

So, we split $3$ as $9 - 6 ,$ and get,

${l}^{2} - \left(9 - 6\right) l - 54 = 0.$

$\Rightarrow \underline{{l}^{2} - 9 l} + \underline{6 l - 54} = 0.$

$\Rightarrow l \left(l - 9\right) + 6 \left(l - 9\right) = 0.$

$\Rightarrow \left(l - 9\right) \left(l + 6\right) = 0.$

$\Rightarrow \left(l - 9\right) = 0 , \mathmr{and} , \left(l + 6\right) = 0.$

$\Rightarrow l = 9 , \mathmr{and} , l = - 6.$

Jun 3, 2017

$l = 9 , \mathmr{and} - 6$

Explanation:

There are 3 ways to solve this problem (by factoring, by completing the square and by using the quadratic formula). I'll use the quadratic formula since the other contributor solved it by factoring.

$54 = {l}^{2} - 3 l$

${l}^{2} - 3 l - 54 = 0$

$a = 1 , b = - 3 , c = - 54$

Substitute it into the quadratic formula which is this

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

$\frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - \left(4\right) \left(1\right) \left(- 54\right)}}{\left(2\right) \left(1\right)}$

Simplify

$\frac{3 \pm \sqrt{225}}{2}$

$\frac{3 + 15}{2} , \frac{3 - 15}{2}$

$\frac{18}{2} , - \frac{12}{2}$

$l = 9 , \mathmr{and} - 6$