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

3 Answers
Jun 3, 2017

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

Jun 3, 2017

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.

Jun 3, 2017

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