In the quadratic equation, solve for x

Question

(1/a)x² + (2/b)x + 3/c = 0

Answer 1

We could use the Quadratic formula to get the roots, but I prefer to complete the square:

(1/a)x² + (2/b)x + 3/c = 0

Subtract 3/c and multiply both sides by 1/a:

(x/a)² + (2/b)(x/a) = -3/ac

Let y = x/a:

y² + (2/b)y = -3/ac

Now complete the square by adding (1/b)² to both sides:

y² + (2/b)y + (1/b)² = (1/b)² - 3/ac

The left side is, by construction, a perfect square:

(y + 1/b)² = (1/b)² - 3/ac

Take the square root of both sides:

y + 1/b = ±sqrt((1/b)² - 3/ac)

Solving for y yields:

y = -1/b ± sqrt((1/b)² - 3/ac)

Remembering that y = x/a, get

x = a⋅y = -a/b ± a⋅ sqrt((1/b)² - 3/ac)

or

x = -a/b ± sqrt((a/b)² - 3a/c)

Answer 2

Use the quadratic formula, which you should be familiar with.

Ax² + Bx + C = 0

x = [-B ± √(B² - 4AC)] / (2A)

Now plug in

A = 1/a

B = 2/b

C = 3/c

and simplify, and get your two solutions.