Question

find zeros of the polynomial 6x^2 + x - 12
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Answer 1

Answer:

By basic determinant method,

x = 【-1 +/- √(1-4(-12)(6)】÷2(6)

x= 【-1 +/- √(1+288)】÷12

on fruther solving,

x= 【-1+/- 17.29】÷12

x=16.29/12 or -18.29

x= 1.35 or -1.52

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Answer 2

Answer:

6x²+x-12=0

6x²+9x-8x-12=0

3x(2x+3)-4(2x-3)=0

(2x-3)(3x-4)=0

=> 2x-3=0 and 3x-4=0

=> x=3/2 and 4/3

Therefore the zeros are 3/2, 4/3

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