Question

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

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Answer 1

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

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

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

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

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

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

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

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