Resposta :
[tex]x+x^2=42\\\\x+x^2-42=42-42\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \left(-42\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-1\pm \:13}{2\cdot \:1}\\\\x_1=\frac{-1+13}{2\cdot \:1},\:x_2=\frac{-1-13}{2\cdot \:1}\\\\\frac{12}{2}=6\\\\\frac{14}{2}=-7\\\\\frac{14}{2}=7[/tex]
[tex]x=6,\:x=-7[/tex]
6 ou -7
Usarei Y para expressar esse número oculto.
Y + Y^2 = 42
Y^2 + Y - 42 = 0
a = 1
b = 1
c = - 42
DELTA:
D = b^2 - 4.a.c
D = 1^2 - 4.1.(-42)
D = 1 + 168
D = 169
Fórmula de Bháskhara:
Y = - b +- raiz quadrada de Delta/ 2.a
Y = - 1 +- raiz quadrada de 169/ 2. 1
Y = - 1 + - 13/ 2
Y1 = - 1 + 13/2
Y1 = 12/2 = 6
Y2 = - 1 - 13/2
Y2 = - 14/2 = - 7
Valores de Y: [ - 7, 6 ]