Question

Alguém pode me ajudar?
-x² + x + 12 = 0

Answer 1

$- x^{2} +x+12=0= \Delta= b^{2} -4.a.c= \Delta 1^{2} -4.(-1).12= \Delta=1+48=49 x^{1} =-+1 \sqrt{49}/2=-1-7/2=-8/2=-4  x^{2} =-+1 \sqrt{49} /2=-1+7/2=6/2=3$


S{-4,3}

Answer 2

$-x^2+x+12=0\\\Delta=1^2-4.(-1).12\\\Delta=1+48\\\Delta=49$

$x= \frac{-b\pm \sqrt{\Delta} }{2a}$

$x= \frac{-1\pm 7 }{-2}$

$x'= \frac{-1+7}{-2}$

$\boxed{x'=-3}$

$x''= \frac{-1-7}{-2}$

$\boxed{x''=4}$

$S=\{-3,4\}$