Question

решите квадратное уравнение 12z²+5z-2=0​

Answer 1

Ответ:

$- \frac{ 2 }{ 3 },\ \frac{ 1 }{ 4 }$

Объяснение:

$12t^{2} + 5t - 2 =0\\\\ a=12 ,\ \ b=5 ,\ \ c=-2\\\\ D = b^2 - 4ac = 5^2 - 4\cdot12\cdot( - 2) = 25 + 96 = 121\\\\\sqrt{D} =\sqrt{121} = 11\\\\ t_1=\frac{-b-\sqrt{D}}{2a}=\frac{-5-11}{2\cdot12}=\frac{-16 }{24 }=- \frac{ 2 }{ 3 } \\\\ t_2=\frac{-b+\sqrt{D}}{2a}=\frac{-5+11}{2\cdot12}=\frac{6}{24}= \frac{ 1 }{ 4 }$

Answer 2

Ответ:

-2/3; 1/4

Объяснение:

12z² + 5z - 2 = 0

a = 12; b = 5; c = -2

D = b² - 4ac

D = 5² - 4 * 12 * (-2) = 25 + 96 = 121

√D = √121 = 11

z1 = (-b - √D)/2a

z1 = (-5 - 11)/(2*12) = -16/24 = -2/3

z2 = (-b + √D)/2a

z2 = (-5 + 11)/(2*12) = 6/24 = 1/4