How do you solve the quadratic using the quadratic formula given k^2+5k-6=0k2+5k−6=0?
1 Answer
Sep 29, 2016
k=1k=1
k=-6k=−6
Explanation:
Given -
k^2+5k-6=0k2+5k−6=0
k=[-b+-sqrt(b^2-4ac)]/(2a)k=−b±√b2−4ac2a
k=[-5+-sqrt(5^2-(4*1*-6))]/(2*1)k=−5±√52−(4⋅1⋅−6)2⋅1
k=[-5+-sqrt(25+24)]/(2*1)k=−5±√25+242⋅1
k=[-5+-sqrt(49)]/(2)k=−5±√492
k=[-5+-7]/(2)k=−5±72
k=[-5+7]/(2)=2/2=1k=−5+72=22=1
k=[-5-7]/(2)=(-12)/2=-6k=−5−72=−122=−6
