factorise then solve PLEASE?

c ² + 7 c - 60 = 0

( c + 12 ) ( c - 5 ) = 0

c = - 12 , c = 5

c² + 7c - 60 =0
c² + 7/2c = 60 + (7/2)²
c² + 7/2c = 240/4 + 49/4
(c + 7/2)² = 289/4
c + 7/2 = 17/2

Factors:
= c + 7/2 - 17/2, = c - 10/2, = c - 5
= c + 7/2 + 17/2, = c + 24/2, = c + 12

Roots:
c - 5 = 0, c = 5
c + 12 = 0, c = - 12

Answer: (c - 5)(c + 12) are the factors; c = 5, - 12

Proof (F.O.I.L.):
= (c - 5)(c + 12)
= c² + 12c - 5c - 60
= c² + 7c - 60

Proof (c = 5):
5² + 7(5) = 60
25 + 35 = 60
60 = 60

Proof (c = - 12):
(- 12)² + 7(- 12) = 60
144 - 84 = 60
60 = 60

(c-5)(c+12) = c² +7c - 60 = 0
so c= 5 or c = -12.

(c+5)(c-12) =0
therefore
c= 5 or c = -12

use the quadratic formula if you ever get stuck

(c+12)(c-5) = 0
c = -12
c = 5

c^2 + 7c - 60 = 0

Factor a trinomial

(1c + 12)(1c - 5)

Subproblem 1

1c + 12 = 0

Add -12 to each side of the equation

1c + 12 - 12 = 0 - 12

c = -12


Subproblem 2

1c - 5 = 0

Add 5 to each side of the equation

1c - 5 + 5 = 0 + 5

c = 5


Solution:

c = -12, 5

c^2 + 7c - 60 = 0
c^2 + 12c - 5c - 60 = 0
(c^2 + 12c) - (5c + 60) = 0
c(c + 12) - 5(c + 12) = 0
(c + 12)(c - 5) = 0

c + 12 = 0
c = -12

c - 5 = 0
c = 5

∴ c = -12, 5

c2+7c-60=0 ok ill give it a shot, think i might be able to do this

c(c+7)-60=0
then to solve i think
c=-7 or c/c=1

i like algebra and i got 100% in my last exam - here goes
cSQ+7c=60
c(c+7)=60 then YOU can solve it lol xx