Question

Calcule o valor de x :

Answer 1

$\sqrt{10} ^{2} = x^{2} + (x+2) ^{2}$

$10 = x^{2} + x^{2} +4x +4$

$2 x^{2} +4x -6=0$

Δ=4^2-4*-6*2
Δ=16+48
Δ=64

x=-b+-rais de Δ / 2a

x=-4+-8/4

x'=-4+8/2 = 4/4 = 1 
x"=-4-8/4= -12/4 = -3

(1,-3)