Question

How do you solve `2x^2+9x+9=0`?

Answer 1

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-9+-sqrt(9^2-4xx2xx9))/(2xx2`

`x=(-9+-sqrt(81-72))/(4`

`x=(-9+-sqrt(9))/(4`

`x=(-9+-3)/(4`

`x=(-9+3)/(4`

`x=(-9-3)/(4`

`x=(-6)/(4`

`x=(-12)/(4`

`x= -3/2,-3`

Explanation:

Plug the values into the Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Answer 2

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

Explanation:

`"'splitting' the middle term to give"`

`2x^2+6x+3x+9=0`

`"taking out a common factor from each 'pair'"`

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

`"taking out the common factor " (x+3)`

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

`"equate each factor to zero"`

`x+3=0tox=-3`

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

`color(blue)"As a check"`

`2(-3)^2+9(-3)+9=18-27+9=0rarr" True"`

`2(-3/2)^2+9(-3/2)+9=9/2-27/2+18/2=0`

`rArrx=-3" or " x=-3/2" are the solutions"`