Respuesta :

gmany
[tex]x^2+2x=15\\\\x^2+2\cdot x\cdot1=15\ \ \ |+1^2\\\\\underbrace{x^2+2\cdot x\cdot1+1^2}_{(a+b)^2=a^2+2ab+b^2}=15+1^2\\\\(x+1)^2=16\to x+1=\pm\sqrt{16}\\\\x+1=-4\ \vee\ x+1=4\ \ \ \ |-1\\\\x=-5\ \vee\ x=3[/tex]
maddymcg408

Answer:

3 AND -5

Step-by-step explanation: