Question

x+x×(3x+4)=23
какое ришение​

Answer 1

Ответ:

$x+x(3x+4)=23\\\\x+3x^2+4x=23\\\\3x^2+5x-23=0\\\\D=b^2-4ac=25+276=301\\\\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{-5+\sqrt{301}}{6}\\\\x_2=\frac{-b-\sqrt{D}}{2a}=\frac{-5-\sqrt{301}}{6}$