Question

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

Answer 1

The solutions are `S={-6 , 9}`

Explanation:

Your equation is

`54=l^2-l3`

We rewrite it as

`l^2-3l-54=0`

We solve this by factorising

`(l+6)(l-9)=0`

The solutions are

`l+6=0`, `=>`, `l=-6`

and

`l-9=0`, `=>`, `l=9`

Answer 2

`l=9, or, l=-6.`

Explanation:

`l^2-3l=54`

`rArr l^2-3l-54=0.`

Now, `9xx6=54, and, 9-6=3.`

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

`l^2-(9-6)l-54=0.`

`rArr ul(l^2-9l)+ul(6l-54)=0.`

`rArr l(l-9)+6(l-9)=0.`

`rArr (l-9)(l+6)=0.`

`rArr (l-9)=0, or, (l+6)=0.`

`rArr l=9, or, l=-6.`

Answer 3

`l= 9, or -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^2-3l`

`l^2-3l-54=0`

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

Substitute it into the quadratic formula which is this

`(-b+-sqrt(b^2-4ac))/(2a)`

`(-(-3)+-sqrt((-3)^2-(4)(1)(-54)))/((2)(1))`

Simplify

`(3+-sqrt(225))/2`

`(3+15)/2, (3-15)/2`

`18/2, -12/2`

`l= 9, or -6`