3c^2-30c+75=0 solve please!?

3(c^2 - 10c + 25) = 0
(c - 5)^2 = 0
c = 5

If its 3a+9=0 then you definately subrtact the 9 to the different section, 3a=9, and divide by capacity of three, a=3 If its 10c^2-30C=0, then you definately would desire to do the quadractic formula accessible or pull out the ten so it would be 10(c^2-3c)=0 and circulate from there wish i helped!

c^2 - 10 c + 25 = 0

( c - 5 ) ( c - 5 ) = 0

c = 5

C=5

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You first have to find the factors of +75 that add up to -30 when one of those factors are multiplied by three. The factors of 75 are
1 * 75
3 * 25
and 5 * 15 - This is the winner. So when you factor you get
(3C - 15) (C - 5) = 0

That means that either
3C - 15 = 0 or/and
C - 5 = 0
You can solve the rest.

Solution: Divide through by 3. c^2 - 10c + 25 = 0

(c-5)^2 = 0

Square root of both sides. c-5=0. Add 5 to both sides. c = 5

first, factor out a 3.
so now it turns into
c^2-10c+25=0
so this factors down to (c-5)(c-5)=0 or (c-5)^2=0 so c=5
to check, remember the quadratic formula:
if you have an equation in the form ax^2+bx+c then,
x=[-b+/-root(b^2-4ac)]/2a
so c=[30+/-root(900-4*3*75)]/6
c=[30+/-root(0)]/6
c=30/6
c=5

I hope this helps!

using splitting the middle term method
3c^2 - 15 c -15 c + 75 = 0
3 c (c - 5) - 15 (c - 5) = 0
(3 c - 15) (c - 5) = 0
3 (c - 5) (c - 5) = 0
since 3 is not equal to zero
so (c - 5) ^2 = 0
c = 5

3c^2 - 30c + 75 = 0
3(c^2 - 10c + 25) = 0
c^2 - 5c - 5c + 25 = 0
(c^2 - 5c) - (5c - 25) = 0
c(c - 5) - 5(c - 5) = 0
(c - 5)(c - 5) = 0

c - 5 = 0
c = 5

∴ c = 5

it factorises to 3(c-5)^2=0 so c=5