How do you solve the quadratic using the quadratic formula given #k^2+5k-6=0#?
1 Answer
Sep 29, 2016
#k=1#
#k=-6#
Explanation:
Given -
#k^2+5k-6=0#
#k=[-b+-sqrt(b^2-4ac)]/(2a)#
#k=[-5+-sqrt(5^2-(4*1*-6))]/(2*1)#
#k=[-5+-sqrt(25+24)]/(2*1)#
#k=[-5+-sqrt(49)]/(2)#
#k=[-5+-7]/(2)#
#k=[-5+7]/(2)=2/2=1#
#k=[-5-7]/(2)=(-12)/2=-6#
