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

2 Answers
May 20, 2017

Answer:

#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)#

May 20, 2017

Answer:

#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"#