Question

How do you solve `3x^2+5x=0` using the quadratic formula?

Answer 1

`x=0`
or
`x=-5/3`

Explanation:

Standard quadratic form:
`color(white)("XXXX")ax^2+bx+c=0`

Re-writing the given equation in explicit standard form:
`color(white)("XXXX")3x^2+5x+0=0`
with `a=3`, `b=5`, and `c=0`

The quadratic formula is
`color(white)("XXXX")x = (-b+-sqrt(b^2-4ac))/(2a)`

which, in this case becomes
`color(white)("XXXX")x = (-5+-sqrt(5^2-4(3)(0)))/(2(3)`

`rArr x= 0 or x=-10/6 = -5/3`

Answer 2

The solutions are

`color(blue)(x=0`

`color(blue)(x=-5/3`

Explanation:

The equation `3x^2+5x=0`:
is of the form `color(blue)(ax^2+bx+c=0` where:

`a=3, b=5, c=0`
(the equation lacks a constant term)

The Discriminant is given by:
`Delta=b^2-4*a*c`
`= (5)^2-(4*3*0)`
`= 25 - 0=25`

The solutions are found using the formula
`x=(-b+-sqrtDelta)/(2*a)`

`x = ((-5)+-sqrt(25))/(2*3) = ((-5+-5))/6`

`x=((-5+5))/6, color(blue)(x=0`

`x=((-5-5))/6, x=-10/6 color(blue)(x=-5/3`