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Mathematics 19 Online
OpenStudy (anonymous):

The absolute value of [x^2 + x] = 12

OpenStudy (pdoctor):

(x^2+x)=12 and (x^2+x)=-12 x^2+x-12=0 and x^2+x+12=0 (x+4)(x-3)=0 and \[x=(-b \pm \sqrt{b^2-4ac})/2a\] So x=-4, 3 and the roots of x^2+x=-12 are imaginary So x= -4, 3

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