Question

t^2 + 78/5 t + 42 = 0​

Answer 1

Answer:

t=−3.4594931171578596,−12.14050688284214

explanation:

by applying quadratic equation formula

∆=-b±√(b^2-4ac)/2a

and by subtituting it we get t=−3.4594931171578596,−12.14050688284214

Answer 2

Answer:

$t_1 = \frac{ - 39 + \sqrt{ 471 } }{5} \\ \\ t_2 = \frac{ - 39 - \sqrt{ 471 } }{5}$

Step-by-step explanation:

To solve the Quadratic equation,first simply it by taking LCM

${t}^{2} + \frac{78}{5} t + 42 = 0 \\ \\ \frac{5 {t}^{2} + 78t + 210}{5} = 0 \\ \\ 5 {t}^{2} + 78t + 210 = 5 \times 0 \\ \\ 5 {t}^{2} + 78t + 210 = 0$

On inspection it is clear that it's factors are not possible.So,apply Quadratic formula.

$t_{1,2}= \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ here \: a = 5 \\ b = 78 \\ c = 210 \\ \\ t_{1,2} = \frac{ - 78 ± \sqrt{ {(78)}^{2} - 4 \times 5 \times 210 } }{2 \times 5} \\ \\ t_{1,2} = \frac{ - 78 ± \sqrt{ 6084- 4200 } }{10} \\ \\ t_{1,2} = \frac{ - 78 ± \sqrt{ 1884 } }{10} \\ \\t_{1,2} = \frac{ - 78 ± 2\sqrt{ 471 } }{10} \\ \\ t_1 = \frac{ - 39 + \sqrt{ 471 } }{5} \\ \\ t_2 = \frac{ - 39 - \sqrt{ 471 } }{5}$

Hope it helps you.