Question

Roots of the equation
$9 {t}^{2} + 2t - 5 = 0$

Answer 1

You can check the nature of the roots by finding Discriminant value by formula (=b^2 - 4ac)...

You have to solve this equation by using the formula.... Hope you know it.
Check it in the pic...

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Answer 2

Heya, 


This is your answer.........

9t² + 2t -5 = 0.
Here, a = 9, b = 2 and c = -5.


First, we will check whether it has root or not ....

Discriminant = b²-4ac.
= 2² - 4(9)(-5)
= 4 + 180
= 184.

As D > 0, the equation has distinct roots.

Now, using formula......

$x = \frac{-b +- \sqrt{D}}{2a}$.
$x = \frac{-2 +- \sqrt{184}}{2 X 9} \\ x = \frac{-2 + \sqrt{184}}{18} or \frac{-2 - \sqrt{184}}{18}$
$x = \frac{-2 + 13.56}{18} or x = \frac{-2 - 13.56}{18} \\ x = \frac{11.56}{18} or x = \frac{-15.56}{18} \\ x = 0.642 or x = -0.864.$

Hence, the roots of above equation are 0.642 and -0.864.

Hope it helps..

If any doubt, feel free to ask....