How do you solve #x^2+x-12=0# using the quadratic formula?

Question

28337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T19:23:10

The answer #x_1=3#

#x_2=-4#

#ax^2+bx+c=0#

#x^2+x-12=0#

#a=1,b=1andc=-12#

#x=[-b+-sqrt(b^2-4ac)]/(2a)#

#x=[-1+-sqrt(1+48)]/(2)#

#x=[-1+-7]/(2)#

#x_1=(-1+7)/2=6/2=3#

#x_2=(-1-7)/2=-8/2=-4#