Question

6t²-19t-7




please tell me the answer​

Answer 1

6t^2-21t+2t-7
3t(2t-7)+1(2t-7)
(2t-7)(3t+1)

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$= > 6 {t}^{2} - 19t - 7 = 0$

=>It is in the form of a quadratic equation $\bold{a {x}^{2} + bx + c}$

We factorise it to find the roots :-

How to factorise..??

step1: Multiply the constant term coefficient of x square.suppose we obtain 'z' after multiplying

step 2:Think of a number whose sum and product makes the 'z'

step 3: now split the middle term by that number which you think.

step 4:At last take common and compare with equal to 0 we obtain the roots .

$= > 6 {t}^{2} - 21t + 2t - 7 = 0$

$= > 3t(2t - 7) + 1(2t - 7)$

$= > (3t + 1)(2t - 7) = 0$

$= > 3t + 1 = 0$

$t = - \frac{1}{3}$

$= > 2t - 7 = 0$

$t = \frac{7}{2}$

Hence ,the roots are :$\bold{\red{t = - \frac{1}{3} }}$ ,$\bold{\red{ t = \frac{7}{2} }}$