Question

Resolva a equação do 2° grau na formula bhaskara X.(X+2)=3

Answer 1

$x(x+2)=3\to x^2+2x-3=0\\\\ a=1; b=2;c=-3\\\\\\ \Delta=b^2-4ac\to \Delta=2^2-4*1*(-3)\to \Delta=4+12\to \Delta=16\\ x' \neq x''\\\\ x= \frac{-b\pm \sqrt{\Delta} }{2a} \to x= \frac{-2\pm \sqrt{16} }{2*1} \to x= \frac{-2\pm 4 }{2} \to \\\\ x'= \frac{-2+ 4 }{2} \to x'= \frac{2 }{2} \to x'=1\\\\ x''= \frac{-2-4 }{2} \to x''= \frac{-6 }{2} \to x''=-3\\\\\\ S=\{-3;1\}$

Answer 2

x.(x + 2) = 3
x² + 2x = 3
x² + 2x - 3 = 0
a = 1; b = 2; c = - 3

Δ = b²  -4ac
Δ = 2² - 4.1.(-3)
Δ = 4 + 12
Δ = 16

x = - b +/- √Δ = - 2 +/- √16
          2a                2.1

x = - 2 + 4 = 2/2 = 1
         2

x = - 2 - 4 = - 6/2 = - 3
          2

R.: x = 1 e x = - 3