Question

1)
$8x { }^{2} + 9x + 1 = 0$
2)
$3x {}^{2} - 4x - 7 = 0$
Пожалуйста решите по Дискриминанту

​

Answer 1

$8x^{2}+9x+1=0$

$D=b^{2}-4ac=9^{2}-4*8*1=49$

$x=\frac{-b-\sqrt{D} }{2a}=\frac{-9-7}{2*8}=-\frac{16}{16}=-1$

$x=\frac{-b+\sqrt{D} }{2a}=\frac{-9+7}{2*8}=-\frac{2}{16}=-\frac{1}{8}$

___________________________

$3x^{2} -4x-7=0$

$D=b^{2}-4ac=(-4)^{2}-4*3*(-7)=100$

$x=\frac{-b-\sqrt{D} }{2a}=\frac{4-10}{2*3}=-\frac{6}{6}=-1$

$x=\frac{-b+\sqrt{D} }{2a}=\frac{4+10}{2*3}=\frac{14}{6}=\frac{7}{3}$