Question

1)4х квадрате+9х+5=0
2)5х квадрате+8х+3=0
помогите плс

Answer 1

$4x^2+9x+5=0$
$D=9^2-4*4*5=81-80= \sqrt{1} = 1$

$x_{1,2} = \frac{-9+-1}{2*4}$

$x_{1}= \frac{-8}{8} = -1$

$x_{2} = \frac{-10}{8} = -1 \frac{2}{8} = -1 \frac{1}{4} = -1,25$

$5x^2+8x+3=0$
$D=8^2-4*5*3=64-60= \sqrt{4} = 2$

$x_{1,2} = \frac{-8+-2}{2*5}$

$x_{1} = \frac{-6}{10} = -0,6$

$x_{2} = \frac{-10}{10} = -1$