Question

How do you solve `8x^{2} + 21= - 59x`?

Answer 1

`x = -7, -0.375`

Explanation:

`8x^2 + 21 = -59 x`

add `59x` to both sides

`8x^2 + 59x + 21`

factor by grouping

`color(white)(.11) + 59`
`color(white)(.11) xx 168`
. . . . . . . . . .
`color(white)(.) color(white)(1) 1 xx 168` `=> 1+168 = 169`
`color(white)(.)color(white)(1) 2 xx 84 color(white)(1)` `=> 2 + 84 = 86`
`color(white)(.)color(white)(1) 3 xx 56 color(white)(1)` `=> 3 + 56 = color(red)(59)`

`(8x^2 + 3x) + (56x + 21)`
`x(8x+3) + 7(8x + 3)`
`(x+7)(8x+3)`
Set each component to `0` and solve for `x`:

`* * * * * * * * * * * * * * * * * * * * * * * * *`

Case 1

`8x + 3 = 0`

`8x = -3`

`x = -3/8`

`* * * * * * * * * * * * * * * * * * * * * * * * *`

Case 2

`x + 7 = 0`

`x = -7`

So `x = -7` and `-3/8` or `-0.375`

Let's graph the equation and see if we are right

graph{y = 8x^2 + 59x + 21}

We are! Nice job

Answer 2

`-3/8, -7`

Explanation:

By factoring:
The factors of 8 are (8,1) or (4,2)
The factors of 21 are (21,1) or (7,3)

`21*8=168` Too large
`21*4=84` Too large
`7*8=56` Worth considering

Rearrange formula to equal zero
`8x^2+59x+21=0`
Since all signs are +, all factors will be +

`(8x+?)(x+?)=0`

By trial-and-error:
`(8x+3)(x+7)=8x^2+56x+3x+21=8x^2+59x+21`

So, `8x+3=0` and `x+7=0` Solve each for `x`

`8x+3=0`
`8x=-3`
`x=-3/8`

And

`x+7=0`
`x=-7`