Question

How do you solve `5r+3=2r^2`?

Answer 1

`r_1=-1/2` and `r_2=3`

Explanation:

`5r+3=2r^2`

`2r^2-5r-3=0`

`(2r+1)*(r-3)=0`

So `r_1=-1/2` and `r_2=3`

Answer 2

`r=-1/2" or "r=3`

Explanation:

`"rearrange into standard form ";ax^2+bx+c=0;a!=0`

`"subtract "5r+3" from both sides"`

`2r^2-5r-3=0`

`"factor using the a-c method"`

`"the factors of the product "2xx-3=-6`

`"which sum to "-5" are "-6" and "+1`

`"split the middle term using these factors"`

`2r^2-6r+r-3=0larrcolor(blue)"factor by grouping"`

`color(red)(2r)(r-3)color(red)(+1)(r-3)=0`

`"take out the "color(blue)"common factor "(r-3)`

`(r-3)(color(red)(2r+1))=0`

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

`2r+1=0rArrr=-1/2`

`r-3=0rArrr=3`