Question

resolva a seguinte equação
${x}^{2} + 2x - 8 = 0$

Answer 1

$x^2+2x-8=0\\\\\Delta=(2)^2-4\cdot(1)\cdot(-8)\\\Delta=4+32\\\Delta=36\\\\x=\frac{-(2)\pm\sqrt36}{2\cdot1}\to x=\frac{-2\pm6}{2}\\\\x'=\frac{-2+6}{2}=2\\\\x''=\frac{-2-6}{2}=-4\\\\s=\{-4,2\}$