Question

hey friends please solve this no tcube-t2square-15t​

Answer 1

Step-by-step explanation:

Given that:

${t}^{3} - {t}^{2} - 15t$

Solution:To solve this polynomial follow these simple steps

Step1: Take t common

$t( {t}^{2} - t - 15) = 0$

Step2:Put both factors equal to zero

$t = 0 \\ \\ {t}^{2} - t - 15 = 0$

Step 3: Now find roots of quadratic equation using quadratic formula

Quadratic formula:$= \frac{- b±\sqrt{ {b}^{2} - 4ac} }{2a}$So, from equation$a = 1 \\ b = - 1 \\ c = - 15 \\ \\ t_{1,2}=\frac{1 ±\sqrt{ {1}^{2} - 4(1)( - 15)} }{2} \\ \\ t_{1,2} = \frac{1 ±\sqrt{61} }{2} \\ \\ t_1 = \frac{1 +7.8}{2} \\ \\ t_1 = \frac{8.8}{2} \\ \\t_1=4.4\\\\ t_2 = \frac{1 - 7.8}{2} \\ \\ t_2 = \frac{ - 6.8}{2} \\ \\t_2=-3.4$

Final Answer: Thus the value of t for the given equation are = 0,4.4,-3.4

or

$0,\frac{1+\sqrt{61} }{2},\frac{1-\sqrt{61} }{2}$

Hope it helps you.