Question

Find the zeroes of the polynomial 5t²+12t+7 *

1 point

–1 , 7/5

1 , –7/5

1/5 , –7

–1 , –7/5​

Answer 1

Answer:

5t^2+12t+7=0

5t^2+5t+7t+7=0

5t(t+1)+7(t+1)=0

t+1=0,5t+7=0

t=-1,5t=-7

t=-1,-7/5

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\sf 5{t}^{2} + 12t + 7 = 0 \\ \\ \sf 5{t}^{2} + 5t + 7t + 7 =0\\ \\ \sf 5t ( t + 1 ) + 7 ( t + 1 ) = 0 \\ \\ \sf ( t + 1 )( 5t + 7 ) =0 \\ \\ \sf So \:the\:zeroes\:are \: -1 \: and \: \frac {-7}{5} \\ \\ \sf \bold{\underline{\underline {So\:choose\:Option\:D}}}$

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