can someone explain how to solve these fraction equatons : c^2/20-c/4+1/5=0?

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can someone explain how to solve these fraction equatons : c^2/20-c/4+1/5=0?

2007-11-04 04:53:57 · 2 answers · asked by Nicole D 1 in Science & Mathematics ➔ Mathematics

2 answers

c^2/20-c/4+1/5=0
Multiply both sides by 20 getting:
c^2 - 5c +4 = 0
(c-1)(c-4) = 0
c = 1 and c = 4

2007-11-04 04:59:07 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋

you can first simply the equation by multiplying it through by 20

c^2/20-c/4+1/5=0 then becomes c^2 - 5c + 4 = 0

then factor it normally

(c-4)(c-1)=0

c = 1 or 4

2007-11-04 04:59:52 · answer #2 · answered by bustedtaillights 4 · 0⤊ 0⤋