Question

How do you solve `3x^2+13x=-4`?

Answer 1

See a solution process below:

Explanation:

First, add `color(red)(4)` to each side of the equation to put the equation in standard form:

`3x^2 + 13x + color(red)(4) = -4 + color(red)(4)`

`3x^2 + 13x + 4 = 0`

Next, we can factor the left side of the equation as:

`(3x + 1)(x + 4) = 0`

Now, solve each term on the left for `0` to find the solutions for `x`:

Solution 1:

`3x + 1 = 0`

`3x + 1 - color(red)(1) = 0 - color(red)(1)`

`3x + 0 = -1`

`3x = -1`

`(3x)/color(red)(3) = -1/color(red)(3)`

`(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -1/3`

`x = -1/3`

Solution 2:

`x + 4 = 0`

`x + 4 - color(red)(4) = 0 - color(red)(4)`

`x + 0 = -4`

`x = -4`

The Solutions Are: `x = -1/3` and `x = -4`

Answer 2

`x=-4" or "x=-1/3`

Explanation:

`"rearrange and equate to zero"`

`rArr3x^2+13x+4=0`

`"the factors of the product of 12 which sum to 13 are"`
`"12 and 1"`

`"split the x-term and factor by grouping"`

`3x^2+12x+x+4=0`

`color(red)(3x)(x+4)color(red)(+1)(x+4)=0`

`"factor out "(x+4)`

`rArr(x+4)(color(red)(3x+1))=0`

`"equate each factor to zero and solve for x"`

`x+4=0rArrx=-4`

`3x+1=0rArrx=-1/3`