Question

Помогите решить пожалуйста 40*x+1*x(x+4)=48*(x+4)

Answer 1

$40 \times x + 1 \times x(x + 4) = 48(x + 4) \\ 40x + x(x + 4) = 48x + 192 \\ 40x + {x}^{2} + 4 x - 48x - 192 = 0 \\ {x}^{2} - 4x - 192 = 0 \\ a = 1 \: \: \: b = - 4 \: \: \: c = 192 \\ d = {b}^{2} - 4ac \\ d = {( - 4)}^{2} - 4 \times 1 \times 192 = 16 + 768 = 784 > 0 \\ x1 = \frac{ 4 - 28}{2} \: \: \: \: x2 = \frac{4 + 28}{2} \\ x1 = - 12 \: \: \: \: \: \: \: \: \: \: x2 = 16$

Answer 2

40*х+1(х+4)=48*(х+4)
40х+х+4=48х+192
41х+4=48х+192
41х-48х=192-4
-7х=188
Х=-26 целых 6/7