Question

2х²+9х+4=0..............​

Answer 1

$\displaystyle 2x^2+9x+4=0\\2x^2+8x+x+4=0\\2x(x+4)+x+4=0\\(x+4)(2x+1)=0\\x+4=0,2x+1=0\\\\ x_{1}=-4,x_{2}=-\frac{1}{2}=-0,5$

Answer 2

$2x {}^{2} + 9x + 4 = 0 \\ a=2~~~~~~~~~b=9~~~~~~~~~~c=4 \\ D=b {}^{2} - 4ac = 9 {}^{2} - 4\times2 \times 4 = 81 - 32 = 49 \\ x1 = \frac{ - 9 + \sqrt{49} }{2 \times 2} = \frac{ - 9 + 7}{4} = \frac{ - 2}{4} = - \frac{1}{2} \\ x2 = \frac{ - 9 - \sqrt{49} }{2 \times 2} = \frac{ - 9- 7}{4} = \frac{ - 16}{4} = - 4 \\ Ответ:~~~~~x1= - \frac{1}{2} ~~~~~~~~~~x2 = - 4$