Question

Resolva a equação, e encontre o delta

${x}^{2}-2x-3 = 0$


​

Answer 1

Resposta:

${x}^{2} - 2x - 3 = 0 \\ {x}^{2} + x - 3x - 3 = 0 \\ x(x + 1) - 3(x + 1) = 0 \\ (x + 1) \times (x - 3) = 0 \\ x + 1 = 0 \\ x - 3 = 0 \\ x = - 1 \\ x = 3$

Answer 2

Resposta:

Explicação passo a passo:

$x^{2} -2x-3=0$

$a=1$

$b=-2$

$c=-3$

$\Delta=b^{2} -4\:.\:a\:.\:c$

$\Delta=(-2)^{2} -4\:.\:(1)\:.\:(-3)$

$\Delta=4+12$

$\Delta=16$

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

$x=\frac{-(-2)\:\pm\:\sqrt{16} }{2.(1)}$

$x=\frac{2\:\pm\:4}{2}$

$x_{1}=\frac{2\:+\:4}{2}=>x_{1} =\frac{6}{2}=>x_{1}=3$

$x_{2}=\frac{2\:-\:4}{2}=>x_{2} =\frac{-2}{2}=>x_{2}=-1$

          $S=\{-1\:;\:3\}$