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

3 Answers
Jun 3, 2017

Answer:

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

Answer:

# 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

Answer:

#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#