How do you solve $k^ { 2} + 23k + 12= - 120$ ?

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Answer 1 · 2018-03-27T17:19:02

$k=-11$
$or$
$k=-12$

$k^2+23k+12=-120$

$k^2+23k+12+120=0$

$k^2+23k+132=0$

$(k+11)(k+12)=0$

$k+11=0$ ==> $k=-11$
$or$
$k+12=0$ ==> $k=-12$

How do you solve k^ { 2} + 23k + 12= - 120k^ { 2} + 23k + 12= - 120?