Question

$x - (x - 32 + 2x^{2} ) = 0$
хэлп плес​

Answer 1

$x - (x - 32 + 2 {x}^{2} ) = 0 \\ x - x + 32 - 2 {x}^{2} = 0 \\ - 2 {x}^{2} + 32 = 0 \\ {x}^{2} - 16 = 0 \\ {x}^{2} = 16 \\ x_{1} = 4 \\ x_{2} = - 4$

Ответ: -4, 4

Answer 2

x-(x-32+2x^2)=0
x-x+32-2x^2=0
2x^2=32
x^2=16
x=4
x=-4