How do you solve #x^2+3x=12# using the quadratic formula?

1 Answer
May 15, 2017

Answer:

#color(blue)(x=2.275# or #color(blue)(x=-5.275# to the nearest 3 decimal places

Explanation:

quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

Quadratic equation: #ax^2 +bx+c =0#

#x^2+3x=12#

#:.x^2+3x-12=0#

#:.a=1,b=3,c=-12#

#:.x=(-(3)+-sqrt((3)^2-4(1)(-12)))/(2(1))#

#:.x=(-3+-sqrt(9+48))/(2(1))#

#:.x=(-3+-sqrt(57))/(2)#

#:.x=(-3+7.549834435)/2# or #x=(-3-7.549834435)/2#

#:.x=(4.549834435)/2# or #x=(-10.549834435)/2#

#:.x=2.2749172175# or #x=-5.2749172175#

#:.color(blue)(x=2.275# or #color(blue)(x=-5.275# to the nearest 3 decimal places

substitute #color(blue)(x=2.2749172175#

#:.(2.2749172175)^2+3(2.2749172175)=12#

#:.5.175248344+6.824751651=12#

#:.12=12#

substitute #color(blue)(x=-5.2749172175#

#:.(-5.2749172175)^2+3(-5.2749172175)=12#

#:.27.82475165-15.82475165=12#

#:.12=12#