Question

How do you solve `4x^2-4x=15`?

Answer 1

`x_1=5/2` or `x_2=-3/2`

Explanation:

Writing your equation in the form
`4x^2-4x-15=0`
dividing by `4`
`x^2-x-15/4=0`
using the quadratic formula
`x_{1,2}=1/2pmsqrt(1/4+15/4)`
so we get

`x_1=5/2`

`x_2=-3/2`

Answer 2

`x=-3/2" or "x=5/2`

Explanation:

`"rearrange in standard form ";ax^2+bx+c=0`

`"subtract 15 from both sides"`

`4x^2-4x-15=0`

`"using the a-c method to factor the quadratic"`

`"the factors of the product "4xx-15=-60`

`"which sum to - 4 are + 6 and - 10"`

`"split the middle term using these factors"`

`4x^2+6x-10x-15=0larrcolor(blue)"factor by grouping"`

`color(red)(2x)(2x+3)color(red)(-5)(2x+3)=0`

`"take out the "color(blue)"common factor "(2x+3)`

`(2x+3)(color(red)(2x-5))=0`

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

`2x+3=0rArrx=-3/2`

`2x-5=0rArrx=5/2`